Question

1 gram of ice is mixed with 1gram of steam. At thermal equilibrium, the temperature of the mixture is
A. $0^{\circ }\text{C}$
B. ${100}^{\circ }\text{C}$
C. ${50}^{\circ }\text{C}$
D. ${55}^{\circ }\text{C}$

Answer 1

Hint: Specific heat capacity of ice, Water and Steam are different. Also we use the concept of latent heat of fusion and vaporization.

Complete step by step solution:
At thermal equilibrium,Total heat gained by ice= total heat lost by steam.
Heat required to convert ice into water,
$Q_1=m_1{L}_f$
where $L_f$ is latent heat of fusion, $m_1$ is the mass of ice.
Substitute $m_1=1\text{g}$ and $80\text{cal}{\text{g}}^{-1}$ to obtain the value of $Q_1$.
$$\begin{gathered} Q_1=1\text{g}\times \text{80}\text{cal}{\text{g}}^{-1} \\ {Q}_1=\text{80}\text{cal} \end{gathered}$$

Now, heat released to convert steam into water,
$Q_2=m_2{L}_v$
where $L_v$ is latent heat of vaporization, $m_2$ is the mass of steam.
Substitute $m_2=1\text{g}$ and $L_v=540\text{cal}{\text{g}}^{-1}$ to obtain the value of $Q_1$.
$$\begin{gathered} Q_2=1\text{g}\times \text{540}\text{cal}{\text{g}}^{-1} \\ {Q}_2=540\text{cal} \end{gathered}$$
It shows that 80 cal is not sufficient to condense steam completely.

Now to convert melted water to ${100}^{\circ }\text{C}$ to $0^{\circ }\text{C}$ is,
$Q=m_{1}S\mathrm{\Delta }T$
Where $S$ is the specific heat of water and $\mathrm{\Delta }T$ is the change in temperature. Therefore,
$$\begin{gathered} Q=1\text{g}\times 1\text{cal}{\text{g}}^{-1}\text{K}\times 100\text{K} \\ Q=100\text{cal} \end{gathered}$$

Therefore, total energy $E$ required to heat ice to water ${100}^{\circ }\text{C}$ is,
$$\begin{gathered} E=100\text{cal + 80}\text{cal} \\ E=180\text{cal} \end{gathered}$$

Thus this amount of energy is also not sufficient for the steam to condense completely. So , the temperature of the mixture is ${100}^{\circ }\text{C}$

Note: The final temperature can also be found by considering the internal energy change of the whole system as constant. Both steam and ice are their respective temperatures to change state.