Question

10g 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

Hint: Water exists in three states: solid (ice), liquid (water), and gas (water) (vapor). Ice is solid water that has been frozen. Water molecules migrate farther apart as it freezes, making ice less solid than water. Liquid water is a kind of water that is both wet and flowing. The vapour of water is still present in the air above us as a gas.
Formula used
Q = mL
where,Q= heat energy
M = per mass released
L = Latent heat of vaporisation

Complete answer: When two physical systems are bound by a heat-permeable path, they are in thermal equilibrium if there is no net transfer of thermal energy between them. If the temperature inside a system is spatially uniform and temporally stable, it is assumed to be in thermal equilibrium with itself.
The final temperature of steam must match that of ice, which is zero, in order to achieve thermal equilibrium.
The heat released by steam as it transitions from gas to liquid is given by Q=mL, where L is the latent heat of vaporisation and m is the mass of the matter.
During a constant-temperature process – typically a first-order phase change – latent heat is emitted or consumed by a body or a thermodynamic device.
Water has a latent heat of evaporation of 540 cal/g and a latent heat of freezing of 80 cal/g, for example.
Given
${{\mathbf{L}}_{{\text{vaporization }}}} = {\mathbf{540cal}}/{\mathbf{g}}$
 m= 10g according to the question
Using Q = m L
${{\text{Q}}_1} = 10 \times 540 = 5400{\text{cal}}$
The heat emitted by boiling water changes its temperature from \[{100^\circ }{\text{C }}to\,{0^\circ }{\text{C}}\]is given as
${\mathbf{Q}} = {\mathbf{mC}}\Delta {\mathbf{T}}\,{\mathbf{cal}}/{\mathbf{g}}$
Where C = The specific heat capacity
$\Delta {\mathbf{T}}\,$ denotes the change in temperature.
${{\text{C}}_{{\text{water }}}} = 1{\text{cal}}/{\text{g}}$
${{\text{Q}}_2} = 10 \times 1 \times (100 - 0) = 1000cal$
Total heating energy transferred = ${{\text{Q}}_1} + {{\text{Q}}_2} = 6400{\text{cal}}$
$6400 = {\text{m}}{{\text{L}}_{{\text{fusion}}}}$
Since latent heat of freezing of 80 cal/g
$6400 = 80m$
m= 80g
$ \Rightarrow {\text{m}} = 80\;{\text{g}}$
And hence option D is the correct answer.

Note:
The energy emitted or consumed by a body or a thermodynamic system during a constant-temperature process — typically a first-order step transition — is known as latent heat (also known as latent energy or heat of transformation).
Latent heat is energy that is supplied or absorbed in a secret way to change the state of a material without changing its temperature. Phase transitions, such as a material condensing or vaporising at a given temperature and pressure, are examples of latent heat of fusion and latent heat of evaporation.