Question

$2g$ steam at ${100}^{\circ }C$ is mixed with $5g$ ice at $-{40}^{\circ }C$ in an ideal calorimeter. The final temperature (in ${}^{\circ }C$) of the system is
$(L_v=500cal/g,s_{ice}=0.5cal/g^{\circ }C,s_w=1cal/g^{\circ }C,L_f=80cal/g)$

Answer 1

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

Given: $m_1=2g,{\theta }_1={100}^{\circ }C,m_2=5g,{\theta }_2=-{40}^{\circ }C,L_v=500cal/g,s_{ice}=0.5cal/g^{\circ }C,s_w=1cal/g^{\circ }C,L_f=80cal/g$

Heat given by steam in converting into water is

$Q_1=m_1{L}_v$

$=2\times 500$

$=1000cal\dots \left(i\right)$

Heat taken by ice to convert to water at ${100}^{\circ }C$ fully is

$Q_2=m_2{s}_{ice}(0-(-40\left)\right)+m_2{L}_f+m_2{s}_w(100-0)$

$Q_2=(5\times 40\times 0.5)+(5\times 80)+(5\times 100\times 1)$

$Q_2=1000cal$

As $Q_1=Q_2$

So, steam and water both are at ${100}^{\circ }C$.

Final answer: ${100}^{\circ }C$.