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Question: 1 gram of ice is mixed with 1gram of steam. At thermal equilibrium, the temperature of the mixture i...

1 gram of ice is mixed with 1gram of steam. At thermal equilibrium, the temperature of the mixture is
A. 0∘C0^\circ {\text{C}}
B. 100∘C100^\circ {\text{C}}
C. 50∘C50^\circ {\text{C}}
D. 55∘C55^\circ {\text{C}}

Explanation

Solution

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,
Q1=m1Lf{Q_1} = {m_1}{L_f}
where Lf{L_f} is latent heat of fusion, m1{m_1} is the mass of ice.
Substitute m1=1  g{m_1} = 1\;{\text{g}} and 80  cal g−180\;{\text{cal}}\,{{\text{g}}^{ - 1}} to obtain the value of Q1{Q_1}.
Q1=1  g×80  cal g−1 Q1=80  cal   {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,
Q2=m2Lv{Q_2} = {m_2}{L_v}
where Lv{L_v} is latent heat of vaporization, m2{m_2} is the mass of steam.
Substitute m2=1  g{m_2} = 1\;{\text{g}} and Lv=540  cal g−1{L_v} = 540\;{\text{cal}}\,{{\text{g}}^{ - 1}} to obtain the value of Q1{Q_1}.
Q2=1  g×540  cal g−1 Q2=540  cal  {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∘  C100^\circ \;{\text{C}} to 0∘C0^\circ {\text{C}} is,
Q=m1SΔTQ = {m_1}S\Delta T
Where SS is the specific heat of water and ΔT\Delta T is the change in temperature. Therefore,
Q=1  g×1  calg−1K×100  K   Q=100  cal  Q = 1\;{\text{g}} \times 1\;{\text{cal}}{{\text{g}}^{ - 1}}{\text{K}} \times 100\;{\text{K}}\; \\\ Q = 100\;{\text{cal}} \\\

Therefore, total energy EE required to heat ice to water 100∘C100^\circ {\text{C}} is,
E=100  cal + 80  cal E=180  cal  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∘C100^\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.