Question

One $\text{gram}$ of ice is mixed with one $\text{gram}$ of steam. After thermal equilibrium is reached, the temperature of the mixture is
(Latent heat of fusion of ice; $L_{ice}=80\text{cal/g}$, latent heat of condensation of vapour; $L_V=540\text{cal/g}$)

• A. ${100}^{\circ }\text{C}$
• B. ${55}^{\circ }\text{C}$
• C. ${50}^{\circ }\text{C}$
• D. $0^{\circ }\text{C}$

Answer 1

The correct option is A ${100}^{\circ }\text{C}$

We know,
Latent heat of fusion of ice; $L_{ice}=80\text{cal/g}$
Latent heat of condensation; $L_V=540\text{cal/g}$ of vapour.
Specific heat capacity of water; $c=1{\text{cal/g}}^{\circ }\text{C}$
Heat supplied by steam if it is to condense totally and convert into water at ${100}^{\circ }\text{C}$.
$Q_1=1\times L_V=540\text{cal}$
Heat required by ice to convert totally into water at $0^{\circ }\text{C}$
$Q_1=1\times L_{ice}=1\times 80=80\text{cal}$
Heat required by ice to convert totally into water at ${100}^{\circ }\text{C}$
$Q_3=1\times L_{ice}+1\times C\times \Delta T=1\times 80+1\times 1\times (100-0)=80+100=180\text{cal}$
Since $Q_1>Q_3$, entire steam will not condense and the final temperature of mixture of water and steam will be ${100}^{\circ }\text{C}$.