Question

An ice cube of mass 0.1 kg at $0$ is placed in an isolated container which is at $227$. The specific heat s of the container varies with temperature T according to the empirical relation s = A + BT, where $A=100cal{kg}^{-1}{K}^{-1}$ and $B=2imes{10}^{-2}cal{kg}^{-1}$ If the final temperature of the container is $27$, the mass of the container is
(Latent heat of fusion of water $=8imes{10}^{4}cal{kg}^{-1}$ , specific heat of water $={10}^{3}cal{kg}^{-1}{K}^{-1}$)

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

Answer 1

The correct option is A 0.495 kg

The temperature at equilibrium is $273+27=300K$

The principle of calorimetry states that the heat lost by container $=$ Heat gained by ice.

Heat lost by the container:

The heat of the container $dQ$ is given as:

$\therefore dQ=mcdT$

Where $C=A+BT$ is specific heat at that temperature

So, $dQ=m(A+BT)dT$

On integrating, we get

$Q{\int }_{500}^{300}m(A+BT)dT=m{[AT+\frac{(BT{)}^2}{2}]}_{500}^{300}$

$=-21600mcalories$

Now. the heat gained by the ice:

For this, the ice is first converted to water and then the energy is used in raising the temperature of the water.

$Q_1=mL=0.1\times 80,000=8000cal$

$Q_2=mc\Delta T=0.1\times {10}^3\times 27=2700cal$

$\therefore Q_1+Q_2=8000+2700=10,700cal$ ...(i)

Heat lost $=$ heat gained

$21600m=10,700$

$\Rightarrow m=0.495kg$