Calorimetry 2j3214

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Heat, Specific Heat and Calorimetry Heat: the basic concept and definition If we place a hot body in with a cold one, the former becomes colder and the latter becomes warmer. Therefore, we can say, a temperature change has occurred between the two bodies and finally a thermal equilibrium is established. We say that a certain quantity of heat has ed from the hot body to the cold one. There is a heat flow or heat transfer between the two bodies. Count Rumford (1753 – 1814) and Sir James Prescott Joule (1818 – 1889) established that this heat flow is an energy transfer. Definition: When an energy transfer takes place due to temperature difference only, it is called the flow of heat. Note: • Heat is energy in transfer due to temperature difference. • Heat is measured in of energy loss or gain and the quantity is described by standard energy units. Quantity of Heat:

We should that the concept of quantity of heat has meaning only in the context of thermal interaction between different systems when energy is transferred from one system to another. A system does not contain a certain quantity of heat; it contains energy; more specifically, internal energy. Unit quantity of Heat: units of heat One unit of heat is defined as the quantity of heat required to raise the temperature of water by 1 0 . Calorie (cal): In the eighteenth century, the unit quantity of heat, the calorie, was defined as the quantity of heat required to raise the temperature of one gram of water by one Celsius degree or one Kelvin. Experimentally, it is found that the heat required to raise the temperature of 1 gm of water by 1 degree is different at different temperatures. For example, the heat required to raise the temperature of 1 gm of water from 0 0 C to 1 0 C is different than the heat required to raise the temperature of the same 1 gm of water from 90 0 C to 91 0 C . So, the required calories are different at different temperatures. Thus the definition of calorie was refined. 15 0 Calorie:

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According to the recommendation of the International Union of Pure and Applied Physics (1934), the standard of Calorie is taken as ‘15 0 Calorie’. The quantity of heat required to raise the temperature of 1 gm of water from 14.5 0 to 15.5 0 at normal atmospheric pressure is called 15 0 Calorie. Mean Calorie: The mean calorie or mean centigrade calorie is defined as the 1/100th of the total quantity of heat required to raise the temperature of 1 gm of water from 0 0 C to 100 0 C . The mean calorie is experimentally found to be 1.0002 times the 15 0 Calorie. Thus the two are essentially the same. Another commonly used unit is kilocalorie as defined below. This is mostly used by biologists or in the case of measuring food energy. Kilocalorie (kcal): The quantity of heat required to raise the temperature of 1 kg of water by 1 0 C is called 1 kilocalorie (kcal). As 1 kg = 1000 gm, we have 1 kcal = 1000 cal. British Thermal Unit (B.Th.U.): In Britain, the frequently used thermal unit is British thermal unit, written as B.th.u. The definition is the following. The quantity of heat required to raise the temperature of 1 pound of water by 1 0 F (from 63 0 F to 64 0 F ) is called 1 B.th.u. Therm: In Britain, the thermal unit used for commercial purpose is ‘Therm’. The quantity of heat required to raise the temperature of 100000 pound of water by 1 0 F is called 1 Therm. 1 Therm = 100000 B.th.u. Note:

Calorie is a thermal unit of C.G.S. system and B.th.u. is of F.P.S. system. There is another unit mixed used in engineering and technological studies. This is called Celsius heat unit (C.H.U.) or Pound calorie. The quantity of heat required to raise the temperature of 1 pound of water by 1 0 C is called 1 Celsius heat unit. Joule and Mechanical equivalent of heat: The English scientist James Joule showed that water can be heated by doing some mechanical work. Joule demonstrated through his experiment that for every 4186 J of work done, the rise in temperature has been 1 0 C for 1 kg of water. This relationship is called the mechanical equivalent of heat.

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Thus the relation between the calorie and joule is 1 cal = 4.186 Joule ≈ 4.2 J. Joule is a thermal unit in S.I. system. The relations among various heat units:

•

1 B.th.u. = 1 pound × 10 F = 453.6 gm ×

•

1 cal =

•

• •

50 C = 252 cal 9

1 B.th.u. = 3.969 × 10 −3 B.th.u. = 4.2 J 252 1 Celsius heat unit = 1 pound × 10 C = 453.6 gm × 1 0 C = 453.6 cal 9 1 Celsius heat unit = 1 pound × 10 C = 1 pound × 0 F = 1.8 B.th.u. 5 5 5 1 Therm = 10 B.th.u. = 252 × 10 cal = 2.52 × 10 7 cal

Heat Transfer: sensible heat

When some quantity of heat is added to a substance, the internal energy of it changes. The heat is used up in increasing the random molecular motion (which results in a temperature change) and also to increase the potential energy associated with the molecular bonds. The sensible heat is referred to that portion of internal energy change in a substance which is associated with a temperature change. Different substances have different molecular arrangements and bonding. Thus for a same amount of heat added to equal mass of different substances, the temperature changes will be different. The quantity of heat, Q that brings a change in temperature, ∆T in a substance, is related by Q ∝ ∆T . Thus we write, Q = C.∆T . The quantity C is called the heat capacity of the substance. In C.G.S unit, C is cal/ 0 C and in S.I. unit, it is J/K. The amount of heat (sensible heat) that is required to change the temperature of a unit mass of a substance is given by: Q = s.∆T , m where m is the mass of the substance and the quantity, s is now specific to the substance, called specific heat capacity or simply specific heat. The term, ‘specific’ relates to the property of unit mass of the substance. Hence we have, Q = m.s.∆T . If we write t = ∆T ,

Q = m.s.t Here, C = m.s . Thus we can write, The heat capacity = mass of the substance × the specific heat.

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Specific Heat: In the expression, Q = m.s.t , if we put m = 1 and t = 1 , we have Q = s . Thus the definition of specific heat is as follows: The specific heat is the amount of heat required to raise the temperature of unit mass of a substance by one degree.

If a system of mass m undergoes a small temperature change dT due to a small heat dQ , 1 dQ we can write, dQ = m.s.dT . Thus the differential form of specific heat is s = . m dT The unit of specific heat in C.G.S. system is cal/gm 0 C or cal gm −1 0 C −1 . In F.P.S. system, it is B.th.u./pound 0 F or B.th.u. pound −1 0 F −1 and in S.I. system, J/kg K or J kg −1 K −1 . The specific heat is thus the number of heat units (as for example, number of calories in C.G.S. system) required to raise the temperature of unit mass of a substance by one degree. We know, 1 unit of heat = the heat required to raise the temperature of unit mass of water by one degree. Thus an alternative definition of specific heat is often given in the following way: The specific heat is the ratio of heat required to raise the temperature of unit mass of a substance by one degree and the heat required to raise the temperature of unit mass of water by one degree.

According to the above definition, specific heat is a dimensionless quantity. This is a number. For example, if we say that the specific heat of copper is 0.09, we understand that the heat required to raise the temperature of unit mass of copper by one degree is 0.09 times the heat required to raise the temperature of unit mass of water by one degree. Note:

• • • •

Specific heat capacity is the heat capacity per unit mass. If M is the molecular weight of a substance, the molar specific heat is Ms . Specific heat of a substance may be considered constant over usual temperature intervals. The greater the specific heat of a substance, the more energy must be transferred to it or taken from it to change the temperature of a given mass of it.

Water Equivalent: Suppose, Q calories of heat raises the temperature of m gm of some substance by t 0 C . We have, Q = m.s.t , where s is the specific heat of the substance. If this Q calories of heat raises the temperature of W gm of water by t 0 C , we can write, Q = m.s.t = W ×1 × t ∴ W = m.s Therefore, we can write the water equivalent of m gm of the substance is W = m.s gm.

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The definition of water equivalent can be given as follows: The quantity of heat that raises the temperature of some substance by some amount, the same quantity of heat that can raise the same temperature of a certain mass of water; the mass of water is then called water equivalent. The heat absorbed or lost by a substance = water equivalent of the substance × temperature increase or decrease. For example, if we say the water equivalent of a substance is 10 gm, we understand that the heat that is required to raise the temperature of the substance by 1 0 C can be used to raise the temperature of 10 gm of water by 1 0 C . Note:

• •

The heat capacity and water equivalent both are equal to the product of the mass of a substance and specific heat. Thus the numerical values of them are equal. But the units are different. By heat capacity we mean some amount of heat whereas the water equivalent is equal to some mass of water. Table # 1: Values of specific heat for some solids and liquids Solids

Lead Brass Copper Aluminium Iron Glass Ice

s cal/gm 0 C

Liquids

s cal/gm 0 C

0.03 0.092 0.093 0.215 0.117 0.16 0.51

Mercury Kerosene oil Tarpin oil Olive oil Alcohol Water

0.033 0.045-0.05 0.42 0.47 0.6 1

Effect of high specific heat of water: In ordinary calorimetric measurements, the specific heat of water is assumed constant and it is equal to 1. Accurate measurements show that it varies with temperature1.

The specific heat of water is quite high compared to other solids and liquids (see table). Thus to raise the same temperature, the water needs more heat than other matters. Also, water needs to lose more heat to cool down to a certain temperature when compared to other solids and liquids. In other words, water takes longer time to get appreciably heated or to be appreciably cooled. For this reason, water is used to keep things warm or cool. 1

The study of the temperature variation of specific heat is the most direct approach to the understanding of molecular energies in matter.

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Water is used in the radiator to keep a car engine cool as water takes long time to get heated. On the other hand, water is used in a hot bag for hot compress because hot water takes long time to cool down. As the value of specific heat of sea water is higher than that of land, sea water does not get much heated throughout the day. The temperature of sea water remains cooler compared to land in the day time. The hot air adjacent to land goes up and the cool air from the sea rushes towards the land, and the sea breeze is created. This does not allow the land to become too hot. On the other hand, the land cools down more easily at night whereas the sea water remains warm for longer time. Then the hot air adjacent to the sea water goes up and the cool air from the land rushes in towards the sea. This way, the places near a sea can neither become excessively cold in winter nor too hot in summer. Problems with Solutions Example 1: The mass of a substance is 200 gm and its specific heat is 0.09. How much heat is required to raise the temperature of the substance from 20 0 C to 90 0 C ? Solution: We know, Q = m.s.t = m.s.(t 2 − t1 ) . Here, m = 200 gm, s = 0.09, t1 = 20 0 C , t 2 = 90 0 C ∴ Q = 200 × 90 × (90 − 20) = 1260 cal. Example 2: What will be the heat capacity and water equivalent of a 200 gm aluminum sheet (specific heat for aluminum = 0.21)? Solution: Here, m = 200 gm, s = 0.21 The heat capacity, C = m.s = 200 × 0.21 = 42 cal/ 0 C and the water equivalent, W = m.s = 200 × 0.21 = 42 gm. Example 3: The ratio of densities of two materials is 2:3 and the ratio of specific heats of them is 0.12:0.09. What will be the ratio of heat capacity of unit volume of the two materials? Solution: Let us suppose, the density of the first material = 2 ρ and the density of the second material = 3ρ ; the specific heat of the first material = 0.12 s and the specific heat of the second material = 0.09 s . The heat capacity of unit volume = the mass of unit volume × specific heat = the density × specific heat If the heat capacities of unit volume for the two materials are H 1 and H 2 , H 1 2 ρ × 0.12 s 2 × 0.12 8 = . = = H 2 3ρ × 0.09 s 3 × 0.09 9

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Example 4: The heat capacities of mercury and glass of equal mass are same. The density of mercury is 13.6 gm/cc and the density of glass is 2.5 gm/cc. If the specific heat of mercury is 0.03, what will be the specific heat of glass? Solution: Let V be the volume of mercury and glass and s be the specific heat of glass. The heat capacity of mercury = V × 13.6 × 0.03 ; the heat capacity of glass = V × 2.5 × s . 13.6 × 0.03 According to question, V × 13.6 × 0.03 = V × 2.5 × s Or, s = = 0.163 . 2.5 Example 5: The water equivalent of a container is 60 gm. The container contains 600 gm water at 30 0 C . How much time will it take to reach the boiling point of water if heat is added to the container at the rate of 100 cal per second? Solution: The heat taken by water = 600 × 1 × (100 − 30) = 600 × 70 = 42000 cal The heat taken by the container = 60 × (100 − 30) = 4200 cal ∴ Total heat taken by the system = 42000 + 4200 = 46200 cal 46200 ∴ The required time = second = 462 sec. = 7 min 42 sec. 100 Example 5: An amount of 10000 B.th.u. of heat can be found from one pound fuel. How much fuel is needed to heat up 50 gallon of water from 45 0 C to 100 0 C ? The mass of 1 gallon of water is 10 pound. Solution: 9 The increase of temperature of water = (100 – 45) = 55 0 C = 55 × = 99 0 F 5 The heat taken by water = 50 × 10 × 1 × 99 = 49500 B.th.u. 49500 ∴ The quantity of required fuel = = 4.95 pound. 10000 Example 6: Find the increase in energy of each aluminum atom if the temperature of a piece of aluminum is increased by 1 0 C . Assume that there are 6 × 10 23 atoms in 27 gm of aluminum and the specific heat of aluminum is 0.2. Solution: The heat required to increase the temperature of 1 gm of aluminum by 1 0 C = 1 × 0.2 × 1 cal = 0.2 × 4.2 J = 0.84 J. [Q 1 cal = 4.2 Joule] 6 × 10 23 Number of atoms in 1 gm of aluminum = . 27 0.84 × 27 ∴ The increase in energy of each atom = = 3.78 × 10 −23 J. 6 × 10 23 Example 7: Show that 0.1 cal/gm 0 C specific heat and 0.1 B.th.u./pound 0 F specific heat are the same. Solution: 5 We know, 1 B.th.u. = 252 cal, 1 pound = 453.6 gm and 1 0 F = 0 C . 9

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0.1 × 252 5 453.6 × 9 0.1 × 252 × 9 = = 0.1 cal/gm 0 C . 453.6 × 5

∴ The specific heat of 0.1 B.th.u./pound 0 F =

Calorimetry: What is calorimetry? Calorimetry is the experimental technique to determine the values of thermal constants, such as specific heat by measuring the quantity of heat. Methods in Calorimetry: The following are some of the methods used in calorimetric measurements. 1. Method of mixing, 2. Method of cooling, 3. Methods based on change of state of matter (e.g., solid to liquid transition) 4. Electrical methods.

We discuss here mainly the method of mixing. Usually, water or some suitable liquid is taken for such calorimetric experiment. The amount of heat is measured by observing the rise in temperature it produces in a known quantity of water or the liquid. Fundamental principle of Calorimetry: When heat is exchanged among different bodies, the heat lost by hot bodies = the heat gained by cold bodies.

Heat is energy. As long as this energy is not converted into some other form, the above principle is applicable and this directly follows from the principle of conservation of energy. What is a calorimeter? In the calorimetric method of mixing, a calorimeter is a specially designed vessel which contains water or some other liquid. Water or the liquid is called the calorimetric substance. Determination of specific heat of a substance:

(By method of mixing)

This study of method of mixing was first done by Regnault in the year 1840. The method consists in taking a substance which is first heated in a furnace up to a high temperature below its melting point. The heat in the substance is then imparted to a certain mass of

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water contained in a vessel of known heat capacity. The quantity of heat is measured by measuring the rise of temperature in water. Let a substance of mass M , specific heat s and initial temperature t1 be dropped into m gm of water at a temperature t 2 . If W be the heat capacity of the calorimeter and t the final temperature of the mixture, we have, the heat lost by the substance = M .s.(t1 − t ) cal; the heat gained by water = m.1.(t − t 2 ) cal and the heat gained by the calorimeter = W (t − t 2 ) cal. [ t 2 = initial temp. of calorimeter] According to the principle of calorimetry, the heat lost = the heat gained. m(t − t 2 ) + W (t − t 2 ) ∴ M .s.(t1 − t ) = m(t − t 2 ) + W (t − t 2 ) Or, s = M (t1 − t ) (m + W )(t − t 2 ) Or, s = M (t1 − t ) Thus we can determine the specific heat of the substance. The above relation can also be used to determine the initial temperature of the substance in case we know the specific heat of it. For precise measurements, various precautions are taken to prevent the loss of heat from the system to surroundings by conduction, convection and radiation. Mixing of liquids:

From calorimetric principle, we know that the total heat lost = total heat gained in a system. In other way, total heat exchange in a system is zero. Let us suppose, three liquids of mass m1 , m2 and m3 are mixed where the specific heat of the liquids are S1 , S 2 and S 3 , respectively and the initial temperatures of the liquids are θ 1 , θ 2 and θ 3 , respectively. If the final temperature of the mixture is θ , we can write, the heat received or lost by the first liquid, Q1 = m1 S1 (θ − θ 1 ) , the heat received or lost by the second liquid, Q2 = m2 S 2 (θ − θ 2 ) , and the heat received or lost by the third liquid, Q3 = m3 S 3 (θ − θ 3 ) . If we consider that there is no heat transferred between the liquid mixture and the surroundings, we can write the total heat exchange among the three liquids is zero: Q1 + Q2 + Q3 = 0 Or, m1 S1 (θ − θ 1 ) + m2 S 2 (θ − θ 2 ) + m3 S 3 (θ − θ 3 ) = 0 m S θ + m2 S 2θ 2 + m3 S 3θ 3 . Or, θ = 1 1 1 m1 S1 + m2 S 2 + m3 S 3 If equal mass of the three liquids are taken ( m1 = m2 = m3 ), we have the final temperature of the liquid mixture, S θ + S 2θ 2 + S 3θ 3 θ= 1 1 . S1 + S 2 + S 3

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Problems with Solutions Example 1: A body of mass 100 gm is heated to 122 0 C and the quickly dropped into a copper calorimeter of mass 50 gm. The calorimeter contains 300 gm water at 28 0 C . The final temperature of the system becomes 30 0 C . What is the specific heat of the body if the specific of copper is 0.09? Solution: Let the specific heat of the solid body = S . The loss of heat by the body = 100 × S × (122 − 30) = 100 × S × 92 = 9200.S cal The gain of heat by water = 300 × 1 × (30 − 28) = 300 × 2 = 600 cal, the gain of heat by the calorimeter = 50 × 0.09 × (30 − 28) = 50 × 0.09 × 2 = 9 cal. ∴ The total heat gained by water and the calorimeter = 600 + 9 = 609 cal. Q The total heat lost = the total heat gained 609 ∴ 9200.S = 609 Or, S = = 0.0662 . 9200 Example 2: A 70 gm solid body is heated and then dropped into a calorimeter containing 116 gm of water. The water equivalent of the calorimeter is 10 gm. If the decrease in temperature of the solid body is 15 times the increase in temperature of water find the specific heat of the solid. Solution: Let the specific heat of the solid body = S and the increase in temperature of the solid body = t 0 C . ∴ The decrease in temperature of the solid body = 15 t 0 C . The heat loss by the solid body = 70 × S × 15.t cal. The heat gained by water = 116 ×1 × t cal and the heat gained by calorimeter = 10 × t cal. ∴ The total heat gained by the calorimeter and water = 116.t + 10.t = 126.t cal. Q The total heat lost = the total heat gained, we can write, 126 70 × S × 15.t = 126.t Or, S = = 0.12 . 70 × 15 Example 3: An alloy contains 60% copper and 40% nickel. A 50 gm piece made of this alloy material is heated up to 50 0 C and plunged into a calorimeter of water equivalent 10 gm where there is 140 gm of water at 20 0 C . What will be the final temperature of the system? (The specific heat for copper = 0.09; the specific heat for nickel = 0.11) Solution: 60 The quantity of copper in the piece of alloy = 50 × = 30 gm and the quantity of 100 nickel = (50 – 30) = 20 gm. Let the final temperature of the system be t 0 C . The heat lost by copper = 30 × 0.09 × (50 − t ) cal and the heat lost by nickel = 20 × 0.11 × (50 − t ) cal. ∴ The total heat lost by the piece of alloy = 30 × 0.09 × (50 − t ) + 20 × 0.11 × (50 − t ) = (50 − t ) × (2.7 + 2.2) = 4.9 × (50 − t ) cal.

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The heat received by the calorimeter = 10 × (t − 20) cal and the heat received by water = 140 × 1 × (t − 20) = 140 × (t − 20) cal. ∴ The total heat gained by water and the calorimeter vessel = 10 × (t − 20) + 140 × (t − 20) = 150 × (t − 20) cal. Q The total heat lost = the total heat gained ∴ 4.9 × (50 − t ) = 150 × (t − 20) Or, 245 − 4.9.t = 150.t − 3000 Or, 154.9.t = 3245 Or, t = 20.95 ∴ The final temperature is 20.95 0 C . Example 4: The water equivalent of calorimeter = 90 gm. If 210 gm of water at 28 0 C is kept for 10 minute, the temperature reduces to 60 0 C . If 100 gm of some other liquid is taken at 80 0 C , the temperature reduces to 60 0 C in 5 minute. If the rate of loss of heat is the same in two cases, what is the specific heat of the liquid? Solution: In the first case, the total loss of heat by water and the calorimeter = 90 × (80 − 60) + 210 × 1 × (80 − 60) = 90 × 20 + 210 × 20 = (90 + 210) × 20 = 300 × 20 = 6000 cal. The loss of heat in 10 minute = 6000 cal 6000 ∴ The rate of loss of heat = = 600 cal/minute 10 Let the specific heat of the liquid = S . In the second case, the total loss of heat by water and the calorimeter = 90 × (80 − 60) + 100 × s × (80 − 60) = 90 × 20 + 100 × s × 20 = 1800 + 2000.S cal. This loss of heat takes 5 minute. 1800 + 2000.S ∴ The rate of loss of heat = cal/minute 5 As the rate of loss of heat is same in the two cases, we can write, 1200 1800 + 2000.S = 600 Or, 1800 + 2000.S = 3000 Or, S = = 0.6 . 2000 5 Example 5: Three liquids A, B and C have temperatures 15 0 C , 25 0 C and 35 0 C . When equal masses of A and B liquids are mixed, the temperature of the mixture becomes 21 0 C . When equal masses of A and B liquids are mixed, the temperature of the mixture becomes 32 0 C . Show that the ratio of specific heats of A and C liquids is 2:7. What will be the temperature of the mixture if A and C liquids are mixed in equal quantity? Solution: Let us assume that the specific heats of A, B and C liquids are S A , S B and S C , respectively. If m unit mass of A is mixed with m unit mass of B, we can say, the heat taken by the liquid A = the heat lost by the liquid B. We can write,

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SA 2 (1) = SB 3 When m unit mass of B is mixed with m unit mass of C, we can write, S 3 m.S B (32 − 25) = m.S C (35 − 32) Or, S B .7 = S C .3 Or, B = (2) SC 7 S S S 2 3 2 ∴ From (1) and (2), we write, A × B = × Or, A = S B SC 3 7 SC 7 ∴ S A : SC = 2 : 7 . m.S A (21 − 15) = m.S B (25 − 21) Or, S A .6 = S B 4 Or,

Now let us assume that the temperature of the mixture of A and C liquids is t 0 C . Considering that the heat received by A-liquid = the heat lost by C-liquid, S 2 35 − t 35 − t m.S A (t − 15) = m.S C (35 − t ) Or, A = Or, = Or, t = 30.56 7 t − 15 S C t − 15

∴ The temperature of the mixture = 30.56 0 C . Example 6: Equal masses of three liquids A, B and C are taken. The temperatures of the liquids are 14 0 C , 24 0 C and 40 0 C . When A and B are mixed, the temperature of the mixture becomes 20 0 C ; when B and C are mixed the temperature becomes 34 0 C . What will be the temperature of the mixture when A, B and C are mixed? Solution: Let the specific heats of A, B and C liquids are S A , S B and S C , respectively. If m unit mass of A is mixed with m unit mass of B, we can say, the heat taken by the liquid A = the heat lost by the liquid B. We can write, S 2 m.S A (20 − 14) = m.S B (24 − 20) Or, S A .6 = S B 4 Or, A = (1) SB 3 When m unit mass of B is mixed with m unit mass of C, we can write, S 3 m.S B (34 − 24) = m.S C (40 − 34) Or, S B .10 = S C .6 Or, B = (2) SC 5 S S S 2 3 2 From (1) and (2) we get, A × B = × Or, A = . S B SC 3 5 SC 5

Now, let us assume that the final temperature of the mixture becomes t 0 C when m unit mass of each of A, B and C liquids are mixed. In this case, the heat lost or received by A is Q A = m.S A (t − 14) unit, the heat lost or received by B is QB = m.S B (t − 24) unit and the heat lost or received by C is QC = m.S C (t − 40) unit. As no heat is transferred with the surroundings, we can write the sum of heat lost or received by the three liquids is zero, Q A + QB + QC = 0 . Thus, m.S A (t − 14) + m.S B (t − 24) + m.S C (t − 40) = 0

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S SA 2 3 (t − 14) + B (t − 24) + (t − 40) = 0 Or, (t − 14) + (t − 24) + (t − 40) = 0 SC SC 3 5 Or, 2.(t − 14) + 3.(t − 24) + 15.(t − 40) = 0 Or, t = 30

Or,

∴ The temperature of the mixture = 30 0 C . Example 7: A 10 gm steel ball of specific heat 0.1 is heated in a furnace and then dropped quickly into a thick copper vessel whose mass is 200 gm and specific heat 0.09 and temperature 50 0 C . Now the whole system is dropped into 180 gm water at 20 0 C contained in a calorimeter of water equivalent 20 gm. The thermometer that is immersed in the calorimeter water shows 26 0 C . What is the temperature of the furnace? Will there be any boiling of water locally? Solution: Let the temperature of the furnace be θ 0 C . The total heat lost by the steel ball and the copper vessel = 10 × 0.1 × (θ − 26) + 200 × 0.09 × (50 − 26) = (θ − 26) + 18 × 24 = θ + 406 cal. The total heat gained by water and the calorimeter container = (20 + 180) × (26 − 20) = 200 × 6 = 1200 cal. Q The total heat lost = the total heat gained ∴ θ + 406 = 1200 Or, θ = 794 ∴ The temperature of the furnace =794 0 C .

Let us now calculate the final temperature when the steel ball is dropped into the copper vessel. If the final temperature of the system is t 0 C , we can write, the heat lost by the steel ball = 10 × 0.1 × (794 − t ) = 794 − t cal and the heat received by the copper vessel = 200 × 0.09 × (t − 50) = 18.t − 900 cal. ∴ 18.t − 900 = 794 − t Or, 19.t = 1694 Or, t = 89.16 . ∴ The final temperature of the system of copper vessel with the steel ball is 89.16 0 C . This temperature is smaller than boiling point of water (100 0 C ). Thus there will not be any local boiling of water. Example 8: Water is flowing through a pipe at a rate of 0.15 kg/minute and it is being heated by a 25.2 watt heater. The temperature of water at inlet is 15.2 0 C and that at outlet is 17.4 0 C . If the flow rate is increased to 0.2318 kg/minute and the rate of heating is increased to 37.8 watt, the temperature of water at inlet and outlet remain unchanged. Find (i) the specific heat of water and (ii) the rate of heat loss from the pipe. Solution: Let the specific heat of water = S J/kg 0 C and the rate of heat loss from the pipe = H watt [1 watt = 1 J/s].

According to question, 0.15 0.15 × S × (17.4 − 15.2) + H = 25.2 Or, × S × 2.2 + H = 25.2 60 60

(1) and

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0.2318 0.2318 × S × (17.4 − 15.2) + H = 37.8 Or, × S × 2.2 + H = 37.8 (2) 60 60 Solving equations (1) and (2) we get, 4200 2.1 S = 4200 J/kg 0 C = cal/gm 0 C = 1 cal/gm 0 C and H = 2.1 J/s = cal/s 4.2 × 1000 4.2 = 0.5 cal/s. Example 9: An empty flask contains 0.3 kg liquid paraffin. If this is heated by a 12.3 watt immersion heater, the temperature of paraffin rises at a rate of 1 0 C every minute. If the same flask contains 0.4 kg liquid paraffin and is heated with a 19.2 watt heater, the temperature of paraffin rises by 1 0 C every minute. Find the specific heat of liquid paraffin and the heat capacity of the flask. Solution: Suppose, the specific heat of liquid paraffin = S and the heat capacity of the flask = C cal/ 0 C . 12.3 × 60 In the first case, the energy produced by 12.3 watt heater in 1 minute = cal. 4.2 So, the total heat taken by paraffin and the flask = 300 × S × 1 + C × 1 = 300.S + C cal. 19.2 × 60 In the second case, the energy produced by 19.2 watt heater in 1 minute = cal. 4.2 So, the total heat taken by paraffin and the flask = 400 × S × 1.2 + C × 1.2 = 480.S + 1.2.C cal. ∴ We can write, 12.3 × 60 300.S + C = (1) and 4.2 19.2 × 60 480.S + 1.2.C = (2) 4.2 Solving (1) and (2) we find, S = 0.529 and C = 17 cal/ 0 C . Example 10: The water equivalents for two calorimeters are 25 gm and 60 gm and both the calorimeters are kept at 0 0 C . Some water of 50 0 C is poured into the first calorimeter and the same water is poured again in the second calorimeter after it reached the final temperature. If the final temperature of water in the second calorimeter is 25 0 C find the quantity of water taken. Solution: Let the quantity of water = m gm and the final temperature of the first calorimeter = θ 0C . In the first calorimeter, the heat lost by water = m × 1 × (50 − θ ) cal; the heat gained by the calorimeter = 25 × θ cal. Q The heat lost = the gained 50.m ∴ m × 1 × (50 − θ ) = 25 × θ Or, θ = (1) 25 + m

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In the second calorimeter, the heat lost by water = m × 1 × (θ − 25) ; the heat gained by the calorimeter = 60 × 25 . 60 × 25 ∴ m × 1 × (θ − 25) = 60 × 25 Or, θ = (2) + 25 m From (1) and (2), 60 50.m 60 × 25 2m + 25 Or, = + 1 Or, 2.m 2 = (m + 25)(m + 60) = 25 + m m 25 + m m Or, (m + 15)(m − 100) = 0 ∴ m = −15 or, 100 (the negative value is not acceptable) ∴ The quantity of water = 100 gm. Example 11: A calorimeter contains 200 gm of water at 10 0 C . A quantity of 50 gm of water at 100 0 C is mixed in it so that the temperature of the mixture now becomes 27 0 C . Next a 100 gm metal ball at 10 0 C is dropped in it and thus the temperature of the system turns to 26 0 C . Find out the specific heat of the metal. Solution: Let the specific heat of the metal = S and the water equivalent of the calorimeter = W gm. In the first case, the heat received by the calorimeter and 200 gm of water in it = the heat lost by 50 gm of added water. 50 × 73 ∴ (W + 200) × 1 × (27 − 10) = 50 × 1 × (100 − 27) Or, W + 200 = (1) 17 Next, the quantity of water in the calorimeter becomes (200 + 50) = 250 gm. Thus in the second case, the heat lost by the calorimeter and 250 gm of water = the heat received by the metal ball. ∴ (W + 250) × 1 × (27 − 26) = 100 × S × (26 − 10) Or, W + 250 = 1600.S (2) Now we get from (2) – (1), 50 × 73 Or, S = 0.165 . 50 = 1600.S − 17 Discussions of a few Questions Q.1 Two kettles, one contains milk and another water of same quantity. If now they are placed on a oven, the temperature of milk is seen to rise than that of water. Explain this. Ans. Let the mass of equal amount of milk and of water is m ; the specific heat of water is S w and the specific heat of milk is S m . As the two similar kettles are placed on the same oven, the rate of heat transfer in them will be the same. Thus the heat Q taken by each of the two kettles will be the same for a certain time. If the rise of temperature in water be θ w and the rise of temperature in milk be θ m , we can write, Q = (W + m.S w )θ w = (W + m.S m )θ m , where W = water equivalent of the kettle.

∴

θ w W + m.S m = θ m W + m.S w

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Now, if we consider, S w > S m , it is possible to have θ m > θ w . Thus the temperature rise in milk will be more than the temperature rise in water for a certain time as the specific heat of water is greater than the specific heat of milk. Q.2 Following Q.1, if we consider that the temperature of milk and water is raised by the same amount, which will receive more heat and why? Ans. If we consider that the rise in temperature in both milk and water is θ , we can write the amounts of heat received by water and milk, respectively Qw = m.S w .θ and Qm = m.S m .θ Q S w > S m ∴ Q w > Qm Thus the water will require more heat than milk for a certain increase in temperature. Q.3 Why 1 pound iron at 100 0 C will be more effective than 1 pound lead at 100 0 C for melting ice? Ans. The specific heat of iron is much more than that of lead. Thus when the temperature of equal mass of iron and lead is brought down from 100 0 C to 0 0 C , the iron will reject more heat than that of lead. Hence 1 pound iron will be more efficient in melting ice. Q.4 What is the advantage of using water in hot bags for the purpose of hot compress? Ans. The specific heat of water is higher than other liquids and solids. For this reason, for the same quantity of supplied heat, the temperature rise in water will be less than any other liquid. Therefore, the heat transfer from the hot water bag to our body will be in a slow rate due to relatively small temperature difference between the hot bag and our body. We thus can be able to have hot compress for a relatively longer time. Q.5 Why is a calorimeter made of metal, particularly of copper instead of any other material e.g., glass? Ans. In a calorimeter, it is required to transfer heat to all parts of it in order to heat up the liquid in an efficient way. The heat conductivity of a metal like copper is much more than that of a material like glass. Thus the liquid in a copper calorimeter comes to thermal equilibrium with the container very quickly and the liquid is heated homogeneously. Also, the specific heat (and so the heat capacity) of copper is less than that of glass. Thus a copper calorimeter takes comparatively less heat to reach a specific temperature. For this reason, it is possible to have a high enough temperature of the calorimetric substance. Thus the relative error in measuring the high temperature will be less. Q.6 A solid copper sphere and another hollow sphere of same size and made of same material are taken. If the two spheres are heated up to the same temperature and then allowed to cool, which one will cool faster? Ans.

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The mass of solid sphere is more than the mass of the similar looking hollow sphere. As the temperatures of the two solid spheres are same, the heat received by the solid sphere is more than the heat received by hollow sphere. The heat to be rejected by a sphere is equal to the heat received by it. Thus the heat rejected by the solid sphere will be more than the heat rejected by the hollow sphere. We can write, the rate of heat rejected = mass × specific heat × rate of decrease in temperature. Thus we can say, being the smaller mass, the hollow sphere will cool down faster. Q.7 Consider the following cases and comment if the fundamental calorimetric principle, ‘heat lost = heat gained’ may be applicable: (i) Sugar is mixed in water in a calorimeter, (ii) there is a chemical reaction between the solid body and the calorimetric liquid, (iii) the calorimeter is kept open (no thermal insulation). Ans. (i) When sugar gets dissolved, it absorbs some amount of heat from water (heat of solution). If this heat is not considered in calculations, the calorimetric principle will not be applicable. (ii) When a chemical reaction occurs, some amount of heat is always either produced (endothermic reaction) or absorbed (exothermic reaction). If this heat is not considered in calculations, we can not apply the calorimetric principle. (iii) If the calorimeter is ‘open’, it exchanges heat with the surroundings. If this heat is not considered into the calculations, the fundamental principle of calorimetry can not be applied. Q.8 What are the advantages of using oil as calorimetric substance over water? Ans. • The specific heat of oil is much less than that of water. Thus for the same amount of supplied heat, the temperature rise in oil is more than that in the same quantity of water. Thus there will be less error involved in measuring temperature rise. • The boiling point of oil is higher than that of water. We should avoid the boiling of the liquid when we do the calorimetric experiment. When a very hot body is dropped in water, it may vaporize and some heat may get lost as the boiling point of water (100 0 C ) is not very high. It is an advantage to work with greater temperature range as in case of oil owing to its high boiling point.

A standard calorimetric substance is aniline. It is found in pure form which has a definite value of low specific heat (0.62) and high boiling point (183.9 0 C ). Questionnaire Very Short Questions:

Mark: 1 (Answer in one or two words)

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1. A piece of wood and a piece of iron are kept at room temperature. Which one of them [iron piece] will appear colder when we touch? [H.S.(XI) ‘06] 2. Which has the highest specific heat among the things that are regularly used? [water] 3. A substance has mass m and specific heat s . How much heat will be needed to increase its temperature by t 0 C ? [ m.s.t ] 4. What is the specific heat of melting ice? [infinity] 5. What is the unit of water equivalent of a body in C.G.S. system? [gram] (Fill in the Blanks)

1. The heat required to raise the temperature of 1 gm water by 1 0 C is called ---------. [calorie] 2. The C.G.S. unit of water equivalent of a substance is ---------------. [gram] 3. The C.G.S. unit of heat capacity of a substance is -----------------. [cal/ 0 C ] 4. The specific heat of water is more than any that of other liquids and solids. Thus water gets heated or cooled with a -------- rate. [slower] 5. The calorie is defined as 1 cal = ------- joule. [4.186] 6. During heat exchange between two bodies, the sum of heat exchange is -----. [zero] (Multiple Choice type)

1. The heat required to raise the temperature of 1 gm of a substance by 1 0 C is called (a) specific heat (b) heat capacity (c) water equivalent (d) latent heat [(a)] 2. The unit of heat in S.I. system is (a) calorie (b) kilocalorie (c) joule (d) watt [(c)] 3. The ‘mean calorie’ is meant by (a) the heat required to increase the temperature of water from 0 0 C to 1 0 C (b) the heat required to increase the temperature of water from 50 0 C to 51 0 C (c) the heat required to increase the temperature of water from 99 0 C to 100 0 C (d) the 1/100 fraction of the heat required to increase the temperature of water from 0 0 C to 100 0 C . [(d)] 4. If Q is the heat required to increase the temperature of m gram of a substance by ∆t , Q m 1 [(b)] (a) ∆t ∝ mQ (b) ∆t ∝ (c) ∆t ∝ (d) ∆t ∝ m Q mQ 5. A substance of m gram requires Q amount of heat for a rise in temperature by t 0 C . If the specific heat of the material of the substance is S , the water equivalent is (a) Q.m (b) m.t (c) m.S (d) S.t [(c)] 0 6. Which of the following has the unit, cal/ gm C ? (a) specific heat (b) heat capacity (c) water equivalent (d) latent heat

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7. Which of the following materials has highest specific heat? (a) mercury (b) water (c) iron (d) diamond

[(b)]

8. The thermal heat capacity of 10 gm of a substance is 8 cal/ 0 C . The specific heat of material of the substance is (a) 0.8 (b) 1.25 (c) 0.4 (d) 0.1 [(a)] 9. If the specific heat of copper 0.1 cal/gm 0 C , the water equivalent of a 0.4 kg copper calorimeter is (a) 40 gm (b) 4000 gm (c) 200 gm (d) 4 gm [(a)] 10. As the specific heat of water is more than that of other liquid or solid (a) water gets heated quickly, but cools slowly (b) water gets heated slowly, but cools quickly (c) water gets heated and cools at a slower rate (d) water gets heated and cools at a faster rate. [(c)] 11. Water is used in the car radiator to keep the engine cool as (a) water is easily available (b) water does not damage the radiator (c) the surface tension of water is much less (d) the specific heat of water is very high. [(d)] 12. The basic principle of calorimetry (heat lost = heat gained) can not be applied when (a) two substances do not mix properly (b) the heat transfer is very high (c) there is a big difference between the specific heats of the two substances (d) there is a chemical reaction between the two substances. [(d)] 13. Which of the following is not normally considered in the calculations of heat lost or heat received? (a) the mass (b) density (c) specific heat (d) temperature change [(b)] 14. Two substances at different temperatures are mixed in a calorimeter. Which one of the following quantities remains constant? (a) Sum of the temperatures of the two substances (b) Total heat of the two substances (c) Sum of the heat capacities of the two substances (d) Sum of the specific capacities of the two substances. [(b)] 15. The ratio of specific heats of two liquids is 1:2. The two liquids of different temperatures are mixed in a mass ratio of 2:3. What will be the ratio of temperature changes in them? (a) 1:3 (b) 1:6 (c) 3:1 (d) 6:1 [(c)] 16. What will be the specific heat of water during boiling? (a) zero (b) 0.5 (c) 1 (d) infinity

[(d)]

17. From which initial value the temperature of 1 gm of water has to be increased so that the value of the received heat will be equal to the mean calorie?

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(a) 0 0 C (b) 14.5 0 C (c) 15 0 C (d) 15.5 0 C

20 [(b)]

18. The alternative of mean calorie is (a) 0 0 C -100 0 C calorie (b) 14.5 0 C calorie (c) 15 0 C calorie (d) 15.5 0 C calorie.

[(c)] 19. Which among the following is the material property of a body? (a) specific heat (b) water equivalent (c) heat capacity (d) temperature Short Questions:

[(a)] Marks: 2

1. On what factors does the received heat by a body depend on? Can we have the concept of specific heat from this? 2. Give the definition of calorie. 3. Give the definition of 15 0 C calorie.

4. Give the definition of mean calorie. 5. Define kilocalorie. 6. What is British thermal unit? 7. What is therm? 8. Give the definition of Celsius heat unit. 9. Give the definition of Fahrenheit heat unit. 10. Define specific heat. 11. What are the units of specific heat in C.G.S., F.P.S. and S.I. systems? [J.E.E. ‘97] 12. What do you mean by specific heat of iron to be 0.106? [H.S. ‘00] [H.S. ’97, ’94, ‘91] 13. Define thermal heat capacity of a substance. 14. Define water equivalent of a substance. [H.S. ’97, ’94, ‘91] 15. What is the difference between heat capacity and water equivalent of a body? [H.S. ’00, ‘98] 16. What are the units of heat capacity and water equivalent in M.K.S. or S.I. unit? [H.S. ‘98] 17. What is the relation between heat capacity and water equivalent of a body? [H.S. ‘99] 18. Show that the heat capacity of unit mass of a substance is equal to specific heat of the material of the substance. [H.S. ‘94] 19. State the fundamental principle of calorimetry and explain this. [H.S. ’02, ’00, ’98, ’96, ’95, ‘92]

20. State the conditions for which the principle of calorimetry is applicable. [H.S. ‘05] 21. For which conditions the principle of calorimetry is not applicable? [H.S. ‘02] 22. What can be measured with a calorimeter? 23. What are the precautions that should be taken during calorimetric measurements? 24. How can the temperature of a furnace be determined by using the calorimetric principle? 25. You are given a thermometer which can measure temperature from 0 0 C to 100 0 C . How can you measure the temperature of a furnace whose temperature is 2000 0 C ? 26. What are the advantages of having high specific heat of water? 27. What is the difference between specific heat and specific gravity? 28. What is the role of specific heat in the atmosphere at sea side?

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29. Two balls of copper and lead of same mass are taken and heated to a same temperature. What will happen if the balls are now placed on a thick paraffin layer? 30. A piece of iron of mass 1 gm at 100 0 C is more effective in melting ice than a piece of lead of same mass and of same temperature – Explain this. [H.S. ‘98] 31. When a cold body is heated by a hot body, do the changes in temperature in them occur at a same rate? Give reasons. 32. The difference in length of a copper rod and a brass rod is same at all temperatures. For what reason is this possible? [H.S. ‘04] 33. Why is water not suitable as a calorimetric substance? [H.S. ‘04] 34. Some liquids of specific heats s1 , s 2 , s 3 ….and at temperatures θ 1 , θ 2 , θ 3 … respectively, are mixed together. What will be the temperature of the mixture? [H.S. ‘05] 35. How is the latent heat for melting of ice determined by applying the principle of calorimetry? [H.S. ‘05] 36. If same heat is supplied to two substances of same mass and different materials, which one of them will be greater temperature? [H.S.(XI) ‘06] 37. Why do we need to thermally insulate a calorimeter from its surroundings? Short Problems:

Marks: 2

1. How much heat is required to heat up a platinum rod from 20 0 C to 70 0 C ? The specific heat of platinum = 0.03 [Ans. 150 cal] 2. The mass of a copper calorimeter is 80 gm. If the specific heat of copper is 0.09, what are the heat capacity and the water equivalent of the calorimeter? [Ans. 7.2 cal, 7.2 gm] 3. The mass of a copper piece is 20 gm. How much will be the temperature rise if 100 cal heat is supplied to it? The specific heat of copper = 0.09 [Ans. 55.5 0 C ] 4. If 30 gm water at 50 0 C is poured into a calorimeter at 15 0 C , the temperature becomes 20 0 C . What is the water equivalent of the calorimeter? [Ans. 180 gm] 5. There is 50 litre air in a room at 30 0 C . The density of air is 1.5 gm/litre. How much heat is required to raise the temperature of air up to 60 0 C ? The specific heat of air = 0.2. [Ans. 450 cal] 0 0 6. A 20 gm substance at 100 C is dropped into 200 gm water at 30 C . What will be the temperature of the system? The specific heat of the substance = 0.1. [Ans. 30.7 0 C ] 7. A steel kettle contains 100 gm of water at 25 0 C . If 50 gm water at 60 0 C is added with this, the temperature of the mixture becomes 35 0 C . Find the water equivalent of the kettle if no heat is lost by radiation or conduction. If the mass of the kettle is 238 gm what is the specific heat of steel? [Ans. 25 gm, 0.105]

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8. The ratio of the densities of two materials is 3:10 and the ratio of the specific heats of them is 7:3. What is the ratio of the thermal heat capacities of the two materials at unit volume? If the ratio of the volumes of the two substances is 1:2, what is their ratio of heat capacities? [Ans. 7:10; 7:20] 9. The mass ratio of two liquids is 3:4, the ratio of their specific heats is 2:3 and their temperatures are 60 0 C and 30 0 C , respectively. What will be the final temperature of the mixture of them? [Ans. 40 0 C ] 10. A platinum ball of mass 80 gm is kept in a furnace in order to measure the temperature of the furnace. When the ball attains the furnace temperature, it is dropped into water at 15 0 C . Now the temperature of water rises to 20 0 C . If the sum of water equivalent of the container and the quantity of water is 400 gm, what is the temperature of the furnace? The specific heat of platinum = 0.0365. [Ans. 704.9 0 C ] 11. The temperatures of two liquids are 80 0 C and 20 0 C . If the two liquids are mixed in a ratio 8:5, the final temperature of the mixture becomes 60 0 C . What is the ratio of specific heats of them? [Ans. 5:4] 12. A liquid of specific heat 0.48 and temperature 25 0 C is mixed with another liquid of specific heat 0.36 and temperature 10 0 C . If the final temperature of the mixture is 20 0 C determine the ratio of the quantities of the two liquids. [Ans. 3:2] 13. In which of the following cases the required heat is greater? (i) The temperature of 500 gm of water is raised from 25 0 C to 100 0 C . (ii) The temperature of 2 pound of water is raised from 82 0 F to 212 0 F .

[Ans. (ii)] 14. The water equivalent of a copper container is 100 gm. This contains 1 kg water at 30 0 C . The container is heated by a Bunsen burner. How long will it take for the water to boil if the burner supplies heat 200 cal per sec? [Ans. 6 m 25 s] 15. The specific heat of a liquid is 0.7 and that of another is 0.4. The water equivalent of 4 litre of the first liquid and the water equivalent of 3 litre of the second liquid are same. [Ans. 3:7] Find the ratio of specific heats of the two liquids. [H.S. ‘04] Harder Problems

1. Three different liquids A, B and C of equal masses are taken. The temperatures of them are 14 0 C , 24 0 C and 40 0 C . When A and B are mixed, the temperature of the mixture becomes 20 0 C ; when B and c are mixed, the temperature of the mixture becomes 34 0 C . What will be the final temperature of the mixture when three liquids are mixed together? [Ans. 30 0 C ]

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2. The temperatures of three liquids A, B and C are 14 0 C , 24 0 C and 34 0 C , respectively. If A and B are mixed in equal masses, the temperature of the mixture becomes 20 0 C ; when B and C are mixed in equal masses, the temperature of the mixture becomes 31 0 C . What will be the final temperature of the mixture if A and C are mixed in equal masses? [Ans. 29.56 0 C ] 3. Three different liquids A, B and C are taken. If 4 gm of liquid A at 60 0 C is mixed with 1 gm of liquid C at 50 0 C , the final temperature of the mixture becomes 55 0 C . When 1 gm of liquid A at 60 0 C is mixed with 1 gm of liquid B at 50 0 C , the final temperature of the mixture becomes 55 0 C . What will be the final temperature of the mixture of 1 gm of liquid B at 60 0 C and 1 gm of liquid C at 50 0 C ? [Ans. 52 0 C ] 4. A liquid in a calorimeter takes 2 min to cool down from 50 0 C to 40 0 C and water of same volume takes 5 min to cool down in the same temperature range in the same calorimeter. The mass of water is 100 gm, the mass of the liquid is 85 gm and the water equivalent of the calorimeter is 10 gm. Determine the specific heat of the liquid. [Ans. 0.4] 5. A vessel contains 200 kg water at 60 0 C . The temperature of water that comes out of a tap is 80 0 C and that from another tap is 20 0 C . Now the two water taps are released over the vessel. If the rate of water flow is 5 kg in 1 min from both the taps, how long does it take for the temperature of water in the vessel to reach 55 0 C ? [Ans. 20 min] 0 6. The mass of a copper vessel is 1 kg and it contains 1 kg of water at 10 C . How much coal is needed to raise the temperature of this water up to boiling point? An amount of 10 4 cal heat is produced by burning 1 kg of coal and 40% of the produced heat is lost. The specific heat of copper = 0.09. [Ans. 16.35 kg] 7. A copper calorimeter with a stirrer weighs 300 gm. The calorimeter contains 200 gm of a liquid up to a certain level. If 41 watt electric power is supplied to the stirrer the temperature increases from 20 0 C to 45 0 C in 10 min. if 140 gm of liquid is withdrawn from the calorimeter and a copper piece of 1250 gm is kept instead, the level of liquid remains unchanged. If the electric power is supplied at the same rate, 2 watt electricity is consumed to keep the temperature fixed for 9 min 5 sec. The room temperature remains constant at 20 0 C during this experiment. Determine the specific heat of copper and that of the liquid. [Ans. 0.0952; 1] 8. A thermometer of mass 55 gm and of specific heat 0.2 shows correct reading at 15 0 C . If this thermometer is immersed into the water in a calorimeter, the reading is 44.4 0 C . The water equivalent of the calorimeter is 50 gm and the mass of water is 250 gm. What was the temperature of water before the thermometer was immersed into it? [Ans. 45.48 0 C ]

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9. The water equivalent of an electric kettle is 100 gm and it contains 800 gm water at 20 0 C . It takes 3 min 45 sec for the water in the kettle to start boiling after the switch is put on. If we assume 20% of heat is lost, find the rate of production of heat in the kettle. [Ans. 440 cal/sec] 10. A vessel whose water equivalent is100 gm, contains 200 gm water at 10 0 C . A heater is supplying heat at a rate of 100 cal/s and 120 gm of water at 5 0 C is being poured into the vessel in every minute. How will it take for the temperature of water to reach [Ans. 2 min] 30 0 C if the vessel and water remain at equal temperature all the time? 11. A 50 gm iron piece and a 50 gm copper piece are soldered together and the system is heated up to 100 0 C . Now it is dropped in 50 gm water at 15 0 C kept in a copper calorimeter and the final temperature becomes 28.35 0 C . When the same process is done taking 100 gm of water, the temperature becomes 22.55 0 C . Find the specific heats iron and copper. [Ans. 0.11; 0.095] 12. A substance of mass M , specific heat S and temperature T 0 is dropped in a liquid of mass m , specific heat s and temperature t 0 . Prove that the final temperature of the MST + mst m(θ − t ) system, θ = . If the liquid is water, show that S = . MS + ms M (T − θ ) 13. A 10 gm copper calorimeter contains 10 gm water at 35 0 C . The temperature is measured by a thermometer. Now this thermometer is replaced by a second thermometer at 15 0 C and then the temperature of water is measured to be 33.35 0 C . Next, the first thermometer is once again inserted and this time it reads 32 0 C whereas the second thermometer reads 32.05 0 C . What is the water equivalent of the second thermometer? (The specific heat of copper = 0.094) [Ans. 1.01 gm] 14. The water equivalent of a vessel is 10 gm. When 100 gm of water at 90 0 C is kept in this uncovered vessel, the temperature reduces to 80 0 C in 5 min. When 50 gm oil is taken, it takes 2 min for the temperature to come down by the same amount. Find the specific heat of oil assuming the same rate of heat loss in the two cases. [Ans. 0.68] 15. If heat is supplied with the help of a 12.3 watt heater to a air-free flask keeping 0.3 kg paraffin oil in it, the temperature rises at the rate of 1 0 C in every minute. If 0.4 kg paraffin oil is taken instead and heat is supplied by 19.2 watt heater, the temperature rises by 1.2 0 C in one minute. Find the specific heat of oil and the thermal heat capacity of the flask. [Ans. 0.529; 17 cal/ 0 C (approx.)] 16. Water and alcohol are taken separately in two similar calorimeters. It is seen that to cool down from 50 0 C to 40 0 C water and alcohol take 100 s and 74 s, respectively. The water equivalent of each of the calorimeters is equal to the numerical value of the volume of respective liquid. Find the specific heat of alcohol. The relative density of alcohol = 0.8. [Ans. 0.6]

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17. A 50 gm metallic alloy substance contains 60% copper and 40% nickel. If this substance is dropped into a calorimeter which contains 55 gm water at 10 0 C , the final temperature becomes 20 0 C . The water equivalent of the calorimeter is 5 gm. Determine the initial temperature of the alloy substance. The specific heat of copper = 0.095 cal/gm 0 C and the specific heat of nickel = 0.11 cal/gm 0 C . [H.S. ‘06] [Ans. 138.8 0 C ]

The End of the Chapter