Question

A calorimeter of water equivalent $10g$ contains a liquid of mass $50g$ 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 $S_{\text{ice }}=0.5cal{g}^{-1}{}^{\circ }{C}^{-1};S_{\text{water }}=1.0cal{g}^{-1}{}^{\circ }{C}^{-1}$ and latent heat of fusion of ice is $L_f=80cal{g}^{-1}$. Assume no heat loss to the surrounding and calculate the value of $m$ in grams.

Answer 1

Answer: 12

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