Question

10 g of steam passes over an ice block. What amount of ice will melt?

• A. 8 g
• B. 18 g
• C. 45 g
• D. 80 g

Answer 1

The correct option is D 80 g

For thermal equilibrium the final temperature of steam must be equal to ice i.e. zero.
The steam will go phase transition from gas to liquid, heat released is given as
$Q=mL$ where L is latent heat of vaporization and m is mass of substance
For water $L_{vaporization}=540cal/g$
$Q_1=10\ast 540=5400cal$
Boiling water will change it's temperature from ${100}^{\circ }C$ to $0^{\circ }C$
heat released is given as
$Q=mC\Delta Tcal/g$ where C is specific heat capacity and $\Delta T$ is temperature difference
$Q_2=10\times 1\times (100-0);C_{water}=1cal/g$
total heat transferred=$Q_1+Q_2=6400cal$
This amount of heat will be used for melting the ice
$6400=m{L}_{fusion}=m80cal/g$ latent heat of fusion for water $=80cal/g$
$\Rightarrow m=80g$