In a laboratory experiment to measure specific heat capacity of copper, 0.05 kg of water at 60°C was poured into a copper calorimeter with a stirrer of mass 0.20 kg initially at 10°C. After stirring, the final temperature reached to 25°C.

Specific heat of water is taken as 4200 J/kg°C.

(a) What is the quantity of heat released per kg of water per 1°C fall in temperature?

(b) Calculate the heat energy released by water in the experiment in cooling from 70°C to 45°C.

(c) Assuming that the heat released by water is entirely used to raise the temperature of calorimeter from 10°C to 45°C, calculate the specific heat capacity of copper.

(a) Quantity of heat released per kg of water per 1°C fall in temperature is given by 4200 J/kg°C.

(b) Given, m_w = 0.05 kg

t_w = 60°C

m_c = 0.20 kg

t_c = 10°C

Specific heat capacity of water = 4200 J kg^-1 K^-1

final temperature of water = 25°C

Heat energy given out by water in lowering it's temperature from 70° C to 45° C
= m x c x ΔT
= 0.05 x 4200 x (70 - 45)
= 0.05 x 4200 x 25
= 5250 J

(c) Heat energy taken by calorimeter when it raises it's temperature from 10°C to 45°C
= m x c x change in temperature
= 0.20 × c × (45 - 10)
= 0.20 × c × 35 = 7c

If there is no loss of energy,

Heat energy gained = heat energy lost

Substituting the values we get,

$5250 = 7c \\[0.5em] \Rightarrow c = \dfrac{5250}{7} \\[0.5em] \Rightarrow c = 750 \text{ J Kg}^{-1}\text{K}^{-1}$

Hence, specific heat capacity of copper = 750 J kg^-1 K^-1