Question

Ice at $0^{\circ }\text{C}$ is added to $1\text{g}$ of steam at ${100}^{\circ }\text{C}$. For $x\text{g}$ of ice added, the temperature of steam reduces to $0^{\circ }\text{C}$. Find the value of $x$.
(Latent heat of ice $=80\text{cal/g}$ and latent heat of steam $=540\text{cal/g}$. Also, specific heat of water $=1\text{cal/ kg K}$)

Answer 1

Given,
Mass of ice $\left(m_{ice}\right)=x\text{g}$
Mass of steam $\left(m_s\right)=1\text{g}$
Initial temperature of steam $\left(T_1\right)={100}^{\circ }\text{C}$
Final temperature of steam $\left(T_2\right)=0^{\circ }\text{C}$
Let $L_f$ be the latent heat of fusion, $L_v$ be the latent heat of vaporization and $c_w$ be the specific heat capacity of water.
For $m\text{grams}$ of ice to just melt, the heat required is given by $Q=m{L}_f$
$\therefore$ For $x\text{grams}$ of ice to just melt, the heat required is given by
$Q_1=(x\times 80)\text{cal}.....\left(1\right)$
Heat required for steam to be condensed and then brought to $0^{\circ }\text{C}$ is given by
$Q_2=m_s{L}_v+m_s{c}_w\Delta T$
$\Rightarrow Q_2=1\times 540+1\times 1\times (100-0)=640\text{cal}$

From principle of calorimetry, we can say that,
Heat lost by steam = Heat gained by ice
$\Rightarrow 640=m\times 80$
$\Rightarrow m=8\text{g}$
$\therefore$ $8\text{g}$ of ice added to $1\text{g}$ of steam reduces the temperature of steam at ${100}^{\circ }\text{C}$ to $0^{\circ }\text{C}$, so that steam condenses and becomes water at $0^{\circ }\text{C}$.