Question

$22320$ cal of heat is supplied to $100$ g of ice at${0^0}{\rm{C}}$. If the latent heat of fusion
of ice is $80{\rm{ cal/g}}$and the latent heat of vaporization of water is $540{\rm{ cal/g}}$, the
final amount of water thus obtained and its temperature respectively are then,
A. $8{\rm{ }}g,{100^0}C$
B. $100{\rm{ }}g,{90^0}C$
C. $92{\rm{ }}g,{\rm{ 10}}{0^0}C$
D. $82{\rm{ }}g,{\rm{ 10}}{0^0}C$

Answer 1

Hint: The latent heat of fusion is the change in its enthalpy due to heat provided to a specific
quantity of a substance to change from its original state like solid to liquid at constant pressure.
The specific heat of water has to be known. The heat required to change from one state to
another is given by, $Q = msd\theta $ where, $Q$is the heat or energy required, $m$is the mass
of the substance, $s$ is the specific heat and $d\theta $is the change in temperature.
Complete step by step solution:
Given,
The latent heat of fusion of ice is, ${L_f} = 80{\rm{ cal/g}}$
The latent heat of vaporization of water is, ${L_v} = 540{\rm{ cal/g}}$
The heat supplied is, $Q = 22320{\rm{ cal}}$
The mass of ice is, $m = 100{\rm{ }}g$
The initial temperature, ${\theta _1} = {0^0}{\rm{ C = 273 K}}$
The specific heat of water, $s = 1{\rm{ cal/g}}{{\rm{m}}^{0{\rm{ }}}}{\rm{C}}$
The equation of latent heat of fusion is given as,
${Q_1} = m{L_f}$ …….(1)
To melt $100{\rm{ g}}$ of ice at ${\rm{273 K}}$ heat required,
${Q_1} = 100 \times 80 = 8000{\rm{ cal}}$
To melt $100{\rm{ g}}$ of water at ${\rm{273 K}}$ to $100{\rm{ g}}$ of water at ${\rm{373
K}}$, heat required,
${Q_2} = msd\theta = 100 \times 1 \times (373 - 273) = 10000{\rm{ cal}}$
To convert $100{\rm{ g}}$ of water at ${\rm{373 K}}$to $100{\rm{ g}}$ of steam at
${\rm{373 K}}$, heat required,
${Q_3} = m{L_v}$ …….(2)
Substituting the values of $100{\rm{ g}}$ in $m$, $1{\rm{ cal/g}}{{\rm{m}}^{0{\rm{
}}}}{\rm{C}}$in $s$and $540{\rm{ cal/g}}$in ${L_v}$
${Q_3} = 100 \times 540 = 54000{\rm{ cal}}$
Thus, total heat required to convert $100{\rm{ g}}$ ice at ${\rm{273 K}}$to $100{\rm{ g}}$
water at ${\rm{373 K}}$ is
$Q = 8000 + 10000 = 18000{\rm{ cal}}$
But to convert the same quantity of ice to steam at ${\rm{373 K}}$, heat required will be
$Q' = 18000 + 54000 = 72000{\rm{ cal}}$
However the supplied heat is given as $22320{\rm{ cal}}$ therefore, all of ice is not converted to
steam. But all of it is necessarily converted to water at 373K first and then some of this water is
converted to steam. The additional heat available after all ice has become water at 373K is
${Q_{additional}} = 22320 - 18000 = 4320{\rm{ cal}}$
Thus if the mass of m g is converted to steam then,
$\begin{array}{l}{Q_{additional}} = m{L_v}\\ \Rightarrow m =
\dfrac{{{Q_{additional}}}}{{{L_v}}}\end{array}$

Substituting the value of $4320{\rm{ cal}}$in ${Q_{additional}}$ and $540{\rm{ cal/g}}$in
${L_v}$,
$m = \dfrac{{4320}}{{540}} = 8{\rm{ g}}$
Thus, from the quantity of $100{\rm{ g}}$ of ice,$92{\rm{ g}}$ is converted to water and
remaining$8{\rm{ g}}$is converted to steam.
373K is nothing but equal to ${\rm{ 10}}{0^0}C$.
Hence the correct answer is (C).
Note: The students have to apply the concept of enthalpy. The relation of heat with mass and
latent heat of both fusion and vaporization has to be known. The heat required to convert ice to
water and water to steam has to be calculated and subtracted from the heat supplied to find the
additional heat. From the additional heat the amount of water can be obtained.