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Question: 60 g of ice at \(0^\circ C\) is mixed with \(60g\) steam at \(100^\circ C\). At thermal equilibrium,...

60 g of ice at 0∘C0^\circ C is mixed with 60g60g steam at 100∘C100^\circ C. At thermal equilibrium, the mixture contains: (Latent heats of steam and ice are 540cal/g540cal/g and 80cal/g80cal/g respectively, the specific heat of water =1cal/g = 1cal/g)
A.) 80g80g of water and 40g40g of steam at 100∘C100^\circ C
B.) 120g120g of water at 90∘C90^\circ C
C.) 120g120g of water at 100∘C100^\circ C
D.) 40g40g of steam and 80g80g of water at 0∘C0^\circ C

Explanation

Solution

Hint: When ice and steam are mixed, first the ice will melt at the melting point through the latent heat of fusion. Then the temperature of melted water will be raised to the temperature of the steam.

Formula used:
Amount of heat energy released during melting is given as
Q=mLQ = mL
where m is the mass of substance while L denotes the latent heat of a substance.

Amount of heat required to change the temperature of a substance from T to T’ is given as
Q=mCΔTQ = mC\Delta T
where C is the specific heat capacity and ΔT\Delta T is the change in temperature.

Detailed step by step solution:
When ice is mixed with steam, first the ice will melt. Melting involves latent heat of fusion so we can find the heat required to melt the ice at its melting point as follows:

Heat required for melting 60 g of ice = Mass of ice x latent heat of ice

Q1=60×80=4800cal{Q_1} = 60 \times 80 = 4800cal

When the ice is melted, the resulting water will undergo an increase in temperature from 0∘C0^\circ C to 100∘C^\circ C. The heat that will be required to increase the temperature is given as

Q2=60×1×100=6000cal{Q_2} = 60 \times 1 \times 100 = 6000cal

Therefore, the total amount of energy required for this whole process is given as

Q=Q1+Q2Q = {Q_1} + {Q_2}

Substituting the known values we get

Q=4800+6000=10800calQ = 4800 + 6000 = 10800cal

Now the steam will also be converted into water because of ice. The amount of steam that will be converted into water will depend on the heat absorbed by the ice and can be calculated as

10800540=20g\dfrac{{10800}}{{540}} = 20g

Hence, the amount of water is equal to the sum of the amount of water due to the melted ice and the amount of steam converted into water.

Therefore, amount of water =60+20=80g = 60 + 20 = 80g

Also the amount of steam has been reduced by 20 g, as this 20 g has been converted into water.

Therefore, amount of steam =60−20=40g = 60 - 20 = 40g

Hence, the correct answer to the question is option A.

Note: The latent heat plays a role only during the change of state. During the change of state there is no increase in temperature. Latent heat of fusion is the heat required to change ice to water at melting point while latent heat of vaporization is the heat required to change water to steam at boiling point.