Question

Ice at ${0^{\text{o}}}{\text{C}}$ is added to ${\text{200}}\,{\text{g}}$ of water initially at ${70^{\text{o}}}{\text{C}}$ in a vacuum flask. When ${\text{50 g}}$ of ice has been added and has all melted the temperature of the flask and contents is ${40^{\text{o}}}{\text{C}}$. When a further ${\text{80 g}}$ of ice has been added and has all melted, the temperature of the whole is ${10^{\text{o}}}{\text{C}}$. Calculate the specific latent heat of fusion of ice.
[ Take ${s_w} = 1\,{\text{cal}}\,{{\text{g}}^{{\text{ - 1}}}}{\,^{\text{o}}}{{\text{C}}^{{\text{ - 1}}}}$ ].
A. $3.8 \times {10^5}\,{\text{J}}\,{\text{k}}{{\text{g}}^{{\text{ - 1}}}}$
B. $1.2 \times {10^5}\,{\text{J}}\,{\text{k}}{{\text{g}}^{{\text{ - 1}}}}$
C. $2.4 \times {10^5}\,{\text{J}}\,{\text{k}}{{\text{g}}^{{\text{ - 1}}}}$
D. $3.0 \times {10^5}\,{\text{J}}\,{\text{k}}{{\text{g}}^{{\text{ - 1}}}}$

Answer 1

Solve the problem in two parts, in the first part find the heat gain and heat loss when ${\text{50 g}}$ of ice is added and then in the second part find the heat gain and heat loss when ${\text{80 g}}$ of ice is added. Form two equations and then find the value of specific latent heat of fusion of ice from these two equations.

Formula used:
The formula for heat transfer is given by,
$\Delta Q = ms\Delta T$ (i)
where $m$ is the mass of the substance, $s$ is the specific heat capacity of the substance and $\Delta T$ is the difference in initial and equilibrium temperature.
The formula for latent heat is,
${Q_L} = mL$ (ii)
where $m$ is the mass of the substance and $L$ is specific latent heat of the substance.

Complete step by step answer:
Given, initial quantity of water, ${m_W} = 200\,g$
Initial temperature of the vacuum flask, ${T_1} = 70{\,^{\text{o}}}{\text{C}}$
Temperature of ice, $T' = 0{\,^{\text{o}}}{\text{C}}$
First quantity of ice added to the water,$m' = 50\,g$
The temperature of the flask after ${\text{50 g}}$ of ice melts, ${T_2} = {40^{\text{o}}}{\text{C}}$
The second quantity of ice added to the water, $m'' = 80\,g$
The temperature of the flask after ${\text{80 g}}$ of ice melts, ${T_3} = {10^{\text{o}}}{\text{C}}$
And specific heat capacity of water, ${s_w} = 1\,{\text{cal}}\,{{\text{g}}^{{\text{ - 1}}}}{\,^{\text{o}}}{{\text{C}}^{{\text{ - 1}}}}$
Let the heat capacity of the flask be $C$ and specific latent heat of fusion of ice be ${L_f}$.We are asked to find specific latent heat of ice fusion of ice.First let us know what specific latent heat of fusion means. Specific latent heat of fusion is the amount of heat required for a substance to convert a unit mass of the substance from its solid state to its liquid state at its melting point.

When ${\text{50 g}}$ of ice is added to the water, the ice will melt and it will take heat from the water and the flask, that is there will be transfer of heat from flask and water to ice. So, the total heat loss will be (using equation (i))
${Q_{loss}} = {m_W}{s_W}({T_1} - {T_2}) + C({T_1} - {T_2})$
Putting the values of ${m_W}$, ${s_W}$, ${T_1}$ and ${T_2}$ we get,
${Q_{loss}} = 200 \times 1 \times (70 - 40) + C(70 - 40)$
$\Rightarrow {Q_{loss}} = 6000 + 30C$ (iii)
Total heat gain by the ice will be latent heat plus the heat gain, (using equation (i) and (ii))
${Q_{gain}} = m'{L_f} + m'{s_W}\left( {{T_2} - T'} \right)$
Putting the values of $m'$, ${s_W}$, $T'$ and ${T_2}$ we get
${Q_{gain}} = 50{L_f} + 50 \times 1 \times \left( {40 - 0} \right)$
$\Rightarrow {Q_{gain}} = 50{L_f} + 2000$ (iv)
Heat loss is equal to heat gain so, equating equations (iii) and (iv) we get,
$6000 + 30C = 50{L_f} + 2000$
$\Rightarrow 600 + 3C = 5{L_f} + 200$
$\Rightarrow 5{L_f} - 3C = 600 - 200$
$\Rightarrow 5{L_f} - 3C = 400$ (v)

Similarly, when ${\text{80 g}}$ of ice has been added further, it takes heat from the water and the flask to melt. Now, the mass of the water will be, ${m_W}^\prime = 200\,{\text{g}} + 50\,{\text{g}}$
So, heat loss by the water and the flask is,
${Q_{loss}} = {m_W}^\prime {s_W}({T_2} - {T_3}) + C({T_2} - {T_3})$
Putting the values of ${m_W}^\prime$, ${s_W}$, ${T_3}$ and ${T_2}$ we get,
${Q_{loss}} = 250 \times 1 \times (40 - 10) + C(40 - 10)$
$\Rightarrow {Q_{loss}} = 7500 + 30C$ (vi)
Total heat gain by the ice will be latent heat plus the heat gain, (using equation (i) and (ii))
${Q_{gain}} = m''{L_f} + m''{s_W}\left( {{T_3} - T'} \right)$
Putting the values of $m''$, ${s_W}$, $T'$ and ${T_3}$ we get
${Q_{gain}} = 80{L_f} + 80 \times 1 \times \left( {10 - 0} \right)$
${Q_{gain}} = 80{L_f} + 800$ (vii)
Heat loss is equal to heat gain so, equating equations (vi) and (vii) we get,
$7500 + 30C = 80{L_f} + 800$
$\Rightarrow 750 + 3C = 8{L_f} + 80$
$\Rightarrow 3C - 8{L_f} = - 670$ (viii)
Now, adding equations (viii) and(v) we get,
$5{L_f} - 3C + 3C - 8{L_f} = 400 - 670$
$\Rightarrow - 3{L_f} = - 270$
$\Rightarrow {L_f} = 90\,{\text{cal}}\,{{\text{g}}^{{\text{ - 1}}}}$
The options are given in joule per kilogram so we convert the unit of specific latent heat of fusion of ice to joule per kilogram.We have,
$1{\text{ cal}} = 4.182\,{\text{J}}$ and ${\text{1 g}} = {10^{ - 3}}{\text{ kg}}$
$\Rightarrow {L_f} = 90{\text{ cal }}{{\text{g}}^{ - 1}} = 90 \times \dfrac{{4.182}}{{{{10}^{ - 3}}}}\,{\text{J k}}{{\text{g}}^{ - 1}}$
$\Rightarrow {L_f} = 376.38 \times {10^3}\;{\text{J k}}{{\text{g}}^{ - 1}}$
$\therefore {L_f} = 3.76 \times {10^5}{\text{ J k}}{{\text{g}}^{ - 1}} \approx 3.8 \times {10^5}{\text{ J k}}{{\text{g}}^{ - 1}}$

Hence, the correct answer is option A.

Note: There are two types of latent heat, which are latent heat of fusion and latent heat of vaporization. Here we have discussed the latent heat of fusion. By latent heat of vaporization we mean the amount of heat required for a liquid substance to change into vapour or gaseous state.