Question

If $120g$ steam of temperature $100{}^{\circ }C$ is released on an ice slab of temperature $0{}^{\circ }C$, how much ice will melt?
(Latent heat of melting of ice = $80cal/g$
Latent heat of vapourization of water = $540cal/g$)

• A. $265g$
• B. $960g$
• C. $691g$
• D. $625g$

Answer 1

The correct option is B $960g$

Given;
Temperature of steam$={\text{T}}_s=100{}^{o}C$
Mass of steam$={\text{m}}_s=120g$
Temperature of ice$={\text{T}}_{ice}=0{}^{o}C$
Latent heat of melting of ice
${\text{L}}_m=80cal/g$
Latent heat of vapourization of water
${\text{L}}_v=540cal/g$

Let heat released during conversion of steam at $100{}^{o}C$ to water at $100{}^{o}C$ be ${\text{Q}}_1$.
$\therefore {\text{Q}}_1={\text{m}}_s\times {\text{L}}_v$
$\therefore {\text{Q}}_1=120\times 540$
$\therefore {\text{Q}}_1=64800cal$

Now, Heat released during conversion of water at $100{}^{o}C$ into water at $0{}^{o}C$ be ${\text{Q}}_2$.
$\therefore {\text{Q}}_2={\text{m}}_s\times \Delta \text{T}\times c$
$\therefore {\text{Q}}_2=120\times (100-0)\times 1$
$\therefore {\text{Q}}_2=12000J$

Therefore, total heat gained by the ice$=\text{Q}={\text{Q}}_1+{\text{Q}}_2=64800+12000$
$=\text{Q}=76800J$

Now, the mass of ice melted due to gain of heat be ${\text{m}}_i$.
$\therefore {\text{m}}_i=\frac{Q}{{\text{L}}_m}$
$m_i=\frac{76800}{80}=960g$