Question

How much steam at ${100}^{\circ }$ will just melt $64gm$ of ice at $-{10}^{\circ }C$?

Answer 1

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

Given: $m_1=64g,{\theta }_1=-{10}^{\circ }C,s_1=0.5cal/g^{\circ }C,s_w=1cal/g^{\circ }C,{\theta }_2={100}^{\circ }C,\theta =0^{\circ }C,L_v=540cal/g,L_f=80cal/g$

Let $m_2$ be mass of steam required.
Heat lost by steam to reach $0^{\circ }C$ will be absorbed by ice during $-{10}^{\circ }C$ to water at $0^{\circ }C$,

$m_2{L}_v+m_2{s}_w(100-0)=m_1\times s_1[0-(-10\left)\right]+m_1\times L_f$

$m_2\times 540+m_2\times 1\times 100=64\times 0.5\times 10+64\times 80$

$640m_2=64\times 85$

$m_2=8.5g$

Final answer: $8.5g$