40g of ice at ${0^ \circ }C$ is used to bring down the temperature of certain mass of water at ${60^ \circ }C$ to ${10^ \circ }C$. Find the mass of water used. (Specific heat capacity of water = 4200 J/kg/C and specific latent heat of fusion of ice = $336 \times {10^3}Jk{g^{ - 1}}$ )

Hint:Whenever heat is added to a system, the temperature of the system increases and vice-versa. However, when there is a transformation in the phase of the object, the heat transfer takes place at constant temperature. This hidden or extra heat is called latent heat.

Complete step-by-step answer:
The matter undergoes phase transformation from one state to another, involving the transfer of heat energy. The change in phase i.e. solid, liquid or gas to another phase happens at a constant temperature depending on the atmospheric pressure. Here, the heat transfer takes place at constant temperature as opposed to normal cases where heat transfer alters the temperature. Hence, there is an excess amount of heat stored or released during phase transformation. This heat is called latent heat.
When 40g of ice is used to extract heat of mass m from ${60^ \circ }C$to ${10^ \circ }C$, the formula for the heat transfer of the water is given by –
$H = mc\Delta T$
c = specific heat capacity of water $\Delta T$ = change in temperature i.e. ${60^ \circ }C$to ${10^ \circ }C$,
The latent heat of 40g of ice is given by the formula –
$L = {m_i}l$
where ${m_i}$= the mass of the ice equal to 40 g or 0.04 kg, $l = $specific latent heat of the substance.
Since the ice is used to cool the water, the latent heat absorbed by the ice is equal to the heat transferred by cooling of water. Therefore,
$H = L$
$mc\Delta T = {m_i}l$
Given,
Specific heat capacity, $c = 4200Jk{g^{ - 1}}{C^{ - 1}}$ Specific latent heat, $l = 336 \times {10^3}Jk{g^{ - 1}}$
Temperature change, $\Delta T = 60 - 10 = {50^ \circ }C$
Substituting the values,
$mc\Delta T = {m_i}l$
$m \times 4200 \times 50 = 0 \cdot 04 \times 336 \times {10^3}$
$ \Rightarrow m = \dfrac{{0 \cdot 04 \times 336 \times {{10}^3}}}{{4200 \times 50}}$
$ \Rightarrow m = \dfrac{{13440}}{{210000}} = 0 \cdot 064kg = 64g$

Hence, the mass of water used is 64 g.

Note:Steam causes severe boils compared to water at boiling point, even though they are at the same temperature, ${100^ \circ }C$. This is because the steam has a higher amount of heat energy than water at boiling point due to the presence of latent heat of evaporation. The value of specific latent heat considered here for ethyl alcohol is actually, the latent heat of evaporation. The latent heat for condensation is the same as that of the specific latent heat for evaporation but, with a negative sign.