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 ${Q_1}$.
$
{Q_1} = 1\;{\text{g}} \times {\text{80}}\;{\text{cal}}\,{{\text{g}}^{ - 1}} \\
{Q_1} = {\text{80}}\;{\text{cal}}\, \\
 $

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 ${Q_1}$.
$
{Q_2} = 1\;{\text{g}} \times {\text{540}}\;{\text{cal}}\,{{\text{g}}^{ - 1}} \\
{Q_2} = 540\;{\text{cal}} \\
$
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,
$Q = {m_1}S\Delta T$
Where $S$ is the specific heat of water and $\Delta T$ is the change in temperature. Therefore,
$
Q = 1\;{\text{g}} \times 1\;{\text{cal}}{{\text{g}}^{ - 1}}{\text{K}} \times 100\;{\text{K}}\; \\
Q = 100\;{\text{cal}} \\
$

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

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.