Question

A calorimeter of water equivalent $10\, g$ contains a liquid of mass $50 \, g$ at $40^{\circ} C$. When $m$ gram of ice at $-10^{\circ} C$ is put into the calorimeter and the mixture is allowed to attain equilibrium, the final temperature was found to be $20^{\circ} C$. It is known that specific heat capacity of the liquid changes with temperature as $S=\left(1+\frac{\theta}{500}\right)$ cal $g ^{-1}{ }^{\circ} C ^{-1}$ where $\theta$ is temperature in ${ }^{\circ} C$. The specific heat capacity of ice, water and the calorimeter remains constant and valuesare $S_{\text {ice }}=0.5\, cal g ^{-1}{ }^{\circ} C ^{-1} ; S_{\text {water }}=1.0 \, cal g ^{-1}{ }^{\circ} C ^{-1}$ and latent heat of fusion of ice is $L_{f}=80 \, cal g ^{-1}$. Assume no heat loss to the surrounding and calculate the value of $m$ in grams.

Answer 1

Heat gained by ice
$=m S_{\text {ice }} \times 10+m S_{m} \times 20 \times m L_{f}=105\, m . cal $
Heat lost by calorimeter $=10 \times 1 \times 20=200 \,cal$
Heat lost by liquid $=-50 \int\limits_{40}^{20}\left(1+\frac{\theta}{500}\right) d \theta$
$=50\left[\theta+\frac{\theta^{2}}{1000}\right]_{20}^{40}=50\left[\left(40+\frac{1600}{1000}\right)-\left(20+\frac{400}{1000}\right)\right] $
$=50 \times 21.2=1060\, cal$
$\therefore 105\, m =1060+200 \Rightarrow m=12 \,g$