Question

$19g$ of water at ${30}^{\circ }C$ and $5g$ of ice at $-{20}^{\circ }C$ are mixed together in a calorimeter. What is the final temperature (in ${}^{\circ }C$) of the mixture? Given specific heat of ice is $0.5cal/g^{\circ }C$ and latent heat of fusion of ice $=80cal/g$.

Answer 1

We know that: $Q=mL,Q=ms\Delta \theta$

Given: $m_1=19g,{\theta }_1={30}^{\circ }C,s_1=1cal/g^{\circ }C,m_2=5g,{\theta }_2=-{20}^{\circ }C,s_2=0.5ca/g^{\circ }C,L=80cal/g$

Let final temperature of mixture be $\theta$.

Heat lost by water is absorbed by ice,
$m_1{s}_1(30-\theta )=m_2{s}_2\left[\right(0-(-20)]+m_{2}L+m_2{s}_1(\theta -0)$

$19\times 1\times (30-\theta )=5\times 0.5\times [0-(-20\left)\right]+5\times 80+5\times 1\times (\theta -0)$

$570-19\theta =50+400+5\theta$

$24\theta =120$

$\theta =5^{\circ }C$

Final answer: $5^{\circ }C$