Question

A calorimeter contains $400g$ of water at a temperature of $5^{\circ }C$. Then, $200g$ of water at a temperature of $+{10}^{\circ }C$ and $400g$ of ice at a temperature of $-{60}^{\circ }C$ are added. What is the final temperature(in ${}^{o}C$) of the contents of the calorimeter? Specific heat capacity of water $=1000cal/kg/K$. Specific latent heat of fusion of ice $=80\times 1000cal/kg$. Relative specific heat of ice $=0.5$

Answer 1

Let $t$ be the common temperature above zero celsius. Heat lost by calorimeter and water added
$=400\times {10}^{-3}\times 1000(5-t)+200\times {10}^{-3}\times 1000(10-t)$
Heat gained by ice$=400\times {10}^{-3}\times 5000(60+t)+400\times {10}^{-3}\times 80000+400\times {10}^{-3}\times 1000t$
Since $Heatlost=heatgained$
$\therefore 400(5-t)+200(10-t)=200(60+t)+400\times 80\times +400t$
or $t=-{24.33}^{\circ }C$
This is contradictory to the assumption that $t$ is above zero. Hence, the final temperature is either $0^{\circ }C$ or less than $0^{\circ }C$. Let the temperature be $t^{\circ }C$ (less than $0^{\circ }C$).
Then $400\times {10}^{-3}\times 1000\times 5+200\times {10}^{-3}\times 1000\times 10+600\times {10}^{-3}\times 80000+600\times {10}^{-3}\times 500\times t=400\times 500(60-t)$
or $400\times 5+200\times 10+600\times 80+300t=200(60-t)$
or $20+20+480+3t=120-2t$
or $t=-{80}^{\circ }C$
This is absurd because the lowest temperature is $-{60}^{\circ }C$ and we discard this second assumption also.
So the final temperature is $0^{\circ }C$.
Let $x$ grams of water be converted into ice.
$Heatlost=400\times {10}^{-3}\times 100\times 5+200\times {10}^{-3}\times 1000\times 10+x\times {10}^{-3}\times 8000=4000+80x$
$Heatgained=400\times {10}^{-3}\times 500\times 60$
or $4000+80x=12000$
or $x=100g$
Hence, the final result is $500g$ of ice and $500g$ of water at $0^{\circ }C$.