What mass of steam at ${\mathbf{100}}^{\mathbf{0}}\mathit{C}$ must be mixed with $\mathbf{\text{150g}}$of ice at its melting point, in a thermally insulated container, to produce liquid water at ${\mathbf{50}}^{\mathbf{0}}\mathit{C}$?

i) The mass of the ice is $m_i=150\text{g}$

ii) The initial temperature of the steam is $T_i=100\text{°C}$

iii) The final temperature of the liquid water is $T_f=50\text{°C}$

The specific heat capacity of the material is the amount of heat required to change the temperature of a unit mass of material by one degree.

We can use the concept of specific heat and heat of transformation. The steam mixes with ice and forms liquid water. Hence, there is a phase change, so we can use the expression of the heat of vaporization. At the same time, the ice is converted to water, so there is the heat of fusion.

Formulae:

The heat energy required by a body$Q=mc\Delta T$ , …(i)

Here, $m$ = mass

$c$= specific heat capacity

$\Delta T$= change in temperature

data-custom-editor="chemistry" $Q$= required heat energy

The heat energy released or absorbed by the body$Q=Lm$ , …(ii)

$L$= specific latent heat

Let $m_s$be the mass of the steam, the energy transferred by the steam when mixed with ice by the expression of specific heat and heat of vaporization is given by using equations (i) and (ii) as:

$Q=c_w{m}_s|T_i-T_f|+L_V{m}_s$

The specific heat of water is$c_w=1\text{\hspace{0.17em}cal/g°C}$

The heat of vaporization is $L_V=539\text{cal/g}$

The initial temperature of the steam is$T_i=100\text{\hspace{0.17em}°C}$, then

$Q=c_w{m}_s|{100}^{0}C-T_f|+L_V{m}_s$ …(iii)

Let $m_i$be the mass of ice, the energy used by the ice when mixed with steam by the expression of specific heat and heat of fusion using equations (i) and (ii) is given by:

$Q=c_w{m}_i(T_f-T_i)+L_F{m}_i$

Where heat of fusion of water is$L_F=79.5\text{cal/g}$

The initial temperature of the ice is $T_i=0.0°\text{C}$, then

$Q=c_w{m}_i(T_f-{0.0}^{0}C)+L_F{m}_i$ …(iv)

When ice is mixed with steam, the energy given by the steam is equal to the energy accepted by the ice, hence, equating equation (iii) with equation (iv) as

$\begin{array}{c}{c}_w{m}_s|100°\text{C}-T_f|+L_V{m}_s=c_w{m}_i(T_f-0.0°\text{C})+L_F{m}_i\\ m_s=\frac{c_w{m}_i(T_f-0.0°\text{C})+L_F{m}_i}{c_w|100°\text{C}-T_f|+L_V}\\ =\frac{1\frac{\text{cal}}{\text{g}°\text{C}}\times 150\text{g}(50°\text{C}-0.0°\text{C})+79.5\frac{\text{cal}}{\text{g}}\times 150\text{\hspace{0.17em}g}}{1\frac{\text{cal}}{\text{g°C}}|100°\text{C}-50°\text{C}|+539\text{cal/g}}\\ =33\text{g}\end{array}$

Hence, the value of the mass of the steam is $33\text{g}$