Calorimetry 2j3214

HEAT AND THERMODYNAMICS

By : Arra C. Quitaneg

Calorimetry - Quantitative measurement of heat exchange. - Employs the “method of mixtures”

- Calorimeter is well insulated so that a minimal amount of heat is exchanged with the outside surroundings.

• If the system is completely isolated, no energy can flow into or out of it. • Energy is conserved • Heat gained = heat loss

Q loss + Q gained = 0

1. If 200g of water at 950C is poured into a 150 g glass cup initially at 250 C, what will be the final temperature of the mixture when equilibrium is reached, assuming no heat flows to the surroundings? C water = 1cal/gC0 C glass = 0.2 cal/gC0

2. We wish to determine the specific heat of a new alloy. A 150 g sample of the alloy is heated to 5400C. It is then quickly placed in 400 g of water at 100C, which is contained in a 200 g aluminum calorimeter cup. The final temperature of the mixture is 30.50C. Calculate the specific heat capacity of the alloy. Cwater = 1cal/gC0 C aluminum = 0.217 cal/gC0

• A 30 g glass thermometer reads 21.6 0 C before it is placed in 115 g of water. When the water and the thermometer come to equilibrium, the thermometer reads 39.20 C. What was the original temperature of the water? Cglass = 0.20 cal/gC0 Cwater = 1 cal/gC0

• When a 290 g piece of iron at 1800 C is placed in a 100 g aluminum calorimeter cup containing 250 g of glycerin at 100 C, the final temperature is observed to be 380 C. What is the specific heat of the glycerin? Ciron = 0.11 cal/gC0 , Caluminum = 0.215 cal/gC0

3. A cube of ice is taken from the freezer at 8.5 0C and placed in a 100 g aluminum calorimeter filled with 300 g of water at temperature of 200C. The final situation is observed to be all water at 170C. What was the mass of the ice cube? Cice = 0.5 cal/gC0 C al = 0.217 cal/gC0 C water = 1 cal/gC0 Lf water = 80cal/g

4. The specific heat of mercury is 138 J/kgC0 . Determine the latent heat of fusion of mercury using the following calorimeter data: 1.00 kg of solid Hg at its melting point of -39.00 C is placed in a 0.620 kg aluminum calorimeter with 0.400 kg of water at 12.8 0 C the resulting equilibrium temperature is 5.06 0 C. Cal = 900 J/kgC0 C water = 4186 J/kgC0

• At a party, a 0.50 kg chunk of ice at -100 C, is placed in 3.0 kg of iced tea at 200 C. At what temperature and in what phase will the final mixture be? The tea can be considered as water. Cwater = 4186 J/kgC0 Ccopper= 390 J/kgC0 Lf ice = 334 x 103 J/kg

A heavy copper pot of mass 2.0 kg is at a temperature of 150 0 C. You pour 0.10 kg of water at 250 C into the pot, then quickly close the lid of the pot so that no steam can escape. Find the final temperature of the pot and its contents. Determine the phase of water, assume no heat is lost to the surroundings. Cwater = 4186 J/kgC0 Ccopper= 390 J/kgC0 Lvwater = 2.256 x 106 J/kg

Seatwork 1. What will be the equilibrium temperature when a 270 g block of copper at 3000C is placed in a 150 g aluminum calorimeter cup containing 820 g of water at 12.0 0C? Cal = 900 J/kgC0 Cwater = 4186 J/kgC0

Ccopper = 390 J/kgC0

Seatwork 2. When 195 g of a substance is heated to 3300 C and then plunged into a 100 g aluminum calorimeter cup containing 150 g of water at 12.5 0 C, the final temperature, is 350 C. What is the specific heat of the substance?

3. The specific heat of mercury is 0.033 kcal/kgC0 . While it is at its melting point of -390 C it is placed in a 0.50 kg aluminum container filled with 1.2 kg of water at 20 0 C. The final temperature of the mixture is found to be 16.50 C. What is the mass of the mercury? LHg = 2.8 kcal/kg

• Suppose that 100 g of ice at an initial temperature of -100 C are put into 400 g of water at an initial temperature of 200 C. How many grams of ice melt. What is the final temperature of the mixture? Lf = 80 cal/g , Cice = 0.5cal/g0