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A thermally insulated vessel contains 150 g of water at \[{{\text{0}}^{\text{0}}}{\text{C}}\]. Then the air from the vessel is pumped out adiabatically. A fraction of water turns into ice and the rest evaporates at \[{{\text{0}}^{\text{0}}}{\text{C}}\] itself. The mass of evaporated water will be closest to: (Latent heat of vaporization of water \[ = 2.0 \times {10^6}J{\text{ }}k{g^{ - 1}}\] and Latent heat of fusion of water \[ = 3.36 \times {10^5}J{\text{ }}k{g^{ - 1}}\] ).
A) $130 g$
B) $35 g$
C) $20 g$
D) $150 g$

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Last updated date: 30th May 2024
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Answer
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Hint: To find the required fraction of water turns into ice we assume m mass of water gets converted into ice and then we will use the principle of the calorimeter and we take the formula: \[{Q_{req}} = m{L_v}\] and \[{Q_{rel}} = m{L_f}\], where \[{Q_{req}}\] and \[{Q_{rel}}\] are the heat required and heat release in the process respectively. m and m’ are the amount of mass of water to ice and rest in the vessel respectively. \[{L_f}\] and \[{L_v}\] are the latent heat of fusion of ice and latent heat of vaporization of water.

Complete step by step answer:
Let ‘m’ mass of water get converted into ice.
To get the required mass of water to be converted into ice formula used: \[{Q_{req}} = m{L_v}\] and \[{Q_{rel}} = m'{L_f}\] here \[{Q_{req}}\] and \[{Q_{rel}}\] are the heat required and heat release in the process respectively. m is the mass of water to be converted into ice. \[{L_f}\] and \[{L_v}\] are the latent heat of fusion of ice and latent heat of vaporization of water.
Given:
The total mass of water = 150 g
\[{L_v} = 2.0 \times {10^6}J{\text{ }}k{g^{ - 1}}\]
\[{L_f} = 3.36 \times {10^5}J{\text{ }}k{g^{ - 1}}\]
If m mass of water converted into ice then the rest amount of water in the container will be = (150 – m)g
Now heat required for evaporation-
\[{Q_{req}} = m{L_v}\]………… (i)
and heat released in the process-
\[{Q_{rel}} = \left( {150 - m} \right){L_f}\]………….(ii)
Applying the principle of calorimeter:
Heat gain = Heat loss
Now,
\[\left( {150 - m} \right){L_f} = m{L_v}\]…………. (iii)
\[ \Rightarrow m\left( {{L_v} + {L_f}} \right) = 150{L_f}\]
\[ \Rightarrow m = \dfrac{{150{L_f}}}{{\left( {{L_v} + {L_f}} \right)}}\]…………….(iv)
Substitute the given value of \[{L_v}\]and \[{L_f}\]eqn (iv), we get
\[ \Rightarrow m = \dfrac{{150 \times 3.36 \times {{10}^5}}}{{\left( {2.0 \times {{10}^6} + 3.36 \times {{10}^5}} \right)}}kg\]
\[ \Rightarrow m = 20{\text{ }}g\]

Hence, option (C) is the correct answer.

Note: In order to master this kind of problem one should have basic understanding of the principle of the calorimeter and its implementation on a given problem. Students often confuse the terms latent heat of vaporization and latent heat of fusion. The latent heat of vaporization is the amount of heat required to convert the unit mass of a substance from liquid to vapor phase while the latent heat of fusion is the amount of heat required to convert the unit mass of a substance from solid to a liquid phase. So it is recommended that students should have to practice a lot of conceptual problems on this topic and build their fundamentals clearly.