Question

A student wishes to determine the heat capacity of a coffee-cup calorimeter. After she mixes 95.8 g of water at 62°C with 95.8 g of water, already in the calorimeter, at 18.2°C, the final temperature of the water is 35.0°C. Calculate the heat capacity of the calorimeter in J/K. Use 4.184 J/g°C as the specific heat of water.

Answer 1

Sol:-

Heat evolved by hot water (q₁) is equal to the sum of heat absorbed by cold water (q₂) and heat absorbed by calorimeter (C).

i.e.

q₁= q₂ + C

C = q₁ - q₂ .............(1)

Heat given evolved by hot water = 95.8 g x 4.184 J/g.⁰C x (35.0 ⁰C - 62.0 ⁰C) = - 10822.3 J
Negative sign indicates that heat is evolved by hot water.

q₂ = 95.8 g x 4.184 J/g.⁰C x (35.0 ⁰C - 18.2 ⁰C) = + 6733.9 J

Positive sign indicates that heat absorbed.

From equation (1), we have
C = 10822.3 J - 6733.9 J = 4088.4 J
ΔT = 35.0 ⁰C - 18.2 ⁰C = 16.8 ⁰C

C = 4088.4 J / 16.8 ⁰C = 243.4 J/^oC