Question

$600\text{g}$ of ice at $253\text{K}$ is mixed with $0.08\text{kg}$ of steam at $100{}^{\circ }\text{C}$. Latent heat of vaporization of steam$=540\text{cal/g}$. Latent heat of fusion of ice $=80\text{cal/g}$. Specific heat of ice $=0.5$${\text{cal/g/}}^{\circ }\text{C}$. The resultant temperature of the mixture is

• A. $273\text{K}$
• B. $300\text{K}$
• C. $330\text{K}$
• D. $373\text{K}$

Answer 1

The correct option is A $273\text{K}$

Given,
Mass of ice, $m_i=600\text{g}$

Mass of steam, $m_s=0.08\text{kg}=80\text{g}$,

Temperature of ice, $T_i=253\text{K}$,

Latent heat of vaporization of steam, $L_v=540\text{cal/g}$,

Latent heat of fusion of ice, $L_f=80\text{cal/g}$,

Specific heat of ice $s_i=0.5{\text{cal/g/}}^{\circ }\text{C}$.

Heat required by $600\text{g}$ of ice at $(253\text{K}=-20{}^{\circ }\text{C})$ to just convert into water at $(273\text{K}=0{}^{\circ }\text{C})$

$Q_1=m_i{s}_i\Delta T+m_i{L}_f$

Substituting the values, we get

$Q_1=600\times 0.5\times (0-(-20\left)\right)+600\times 80$

$Q_1=6000+48000=54000\text{cal}$

The maximum heat released by steam when the whole of its mass $\left(80\text{g}\right)$ is converted into water at $0^{\circ }\text{C}$ is

$Q_2=m_s{s}_s\Delta T+mL$

$\Rightarrow Q_2=80\times 1\times (100-0)+540\times 80$

$\Rightarrow Q_2=8000+43200=51200\text{cal}$

Since $Q_2$ is less than $Q_1$ the whole of ice will not melt. Hence the final temperature of the mixture will be $0^{\circ }\text{C}$ or $273\text{K}$.

Hence, option (a) is the correct answer.