Question

An ice cube of mass 0.1 kg at 0°C is placed in an isolated container which is at 227°C. The specific heat s of the container varies with temperature T according to the empirical relation s = A + BT, where A =100 cal$kg^{- 1}K^{- 1}$and $B = 2 \times 10^{- 2}calkg^{- 1}$ If the final temperature of the container is 27°C, the mass of the container is

(Latent heat of fusion of water $= 8 \times 10^{4}calkg^{- 1}$ specefic heat of water $= 10^{3}Calkg^{- 1}k^{- 1}$)

• A. 0.495 kg
• B. 0.595 kg
• C. 0.695 kg
• D. 0.795 kg

Answer 1

Answer: (A)

0.495 kg

Answer 2

Heat lost by container

$= - \int_{500}^{300}m_{C}$ (A+BT)dT

$= - m_{C}\left\lbrack AT + \frac{BT^{2}}{2} \right\rbrack_{500}^{300} = 21600m_{C}$

Heat gained by ice

$= m_{ice}L + m_{ice}s_{water}\Delta T$

$= 0.1 \times 8 \times 10^{4} + 0.1 \times 10^{3} \times 27$

= 10700 cal

According to principle of calorimetry Heat lost by constainer = Heat gained by ice

21600$m_{c} = 10700$

Or $m_{c} = 0.495kg$