1. A student determined the calorimeter constant of the calorimeter, using the p

Question

1. A student determined the calorimeter constant of the calorimeter, using the procedure described in this module. The student added 50.00 mL of cold water to 50,00 mL of heated distilled water in a Syrfoam cup. The in a Styrofoam cup. The initial temperature of the cold water was 19.09 °C maximum temperature of the mixture was found to be and of the hot water, 32.17 °C. The 24.55 °C. Assume the density of water is 1.00 g mlL Calculate the heat lost by the hot water. and the specific heat is 4.184 J g K 2. Calculate the heat gained by the cold water. 3. Calculate the calorimeter constant using the AT of the cold water. Enthalpy

Explanation / Answer

mass = volume x density

mass of hot water = mass of cold water = 50 ml x 1 g/ml = 50 g

1) heat lost by hot water = mass x specific heat x temp change

heat lost by hot water = m x s x dT

heat lost by hot water = m x s x (Ti - Tf)

heat lost by hot water = 50 x 4.184 x (32.17-24.55)

heat lost by hot water = 1594.104 J

2)

heat gained by cold water = 50 x 4.184 x (24.55 - 19.99)

heat gained by cold water = 953.952 J

3)

heat gained = heat lost

now

heat gained by cold water + heat gained by calorimeter = heat lost by hot water

953.952 + heat gained by calorimeter = 1594.104

heat gained by calorimeter = 640.152 J

now

heat gained by calorimeter = C x dT

640.152 = C x ( 24.55 - 19.99)

C = 140.384 J/ C

so

the calorimeter constant is 140.384 J/ C

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