Given data:
The mass of steam is 100 g.
The mass of water is 200 g.
The mass of ice is 20 g.
The initial temperature of water and ice is 0°C.
The initial temperature of steam is 100°C.
The calorimeter behaves thermally equivalent to 30 g of water.
Formula used:
The heat interaction of a body due to change in temperature is expressed by the formula,
ΔQ=mcΔT
Here, ΔQ is the heat transferred to the body, ΔT is the change in temperature of the body, m is the mass of the body, and c is the specific heat of the body.
The expression for heat of fusion is,
Qf=mLf
Here, Lf is the latent heat of fusion of the body and Qf is the heat of fusion.
The expression for heat of condensation is,
Qv=mLv
Here, m is the mass of steam, Lv is the latent heat of condensation of body, and Qv is the heat required to condense mass m.
The expression for net heat transfer in a calorimeter is written as,
∑Qnet=0
Here, ∑Qnet is the net heat transferred in a calorimeter.
Explanation:
Consider that the final temperature of the mixture is less than 100°C, that is, all the steam condenses into water and the heat that is released by the steam is gained by the water-and-ice mixture initially at 0°C.
Calculate the heat gained by steam to condense into water.
Qv1=m1Lv1
Here, m1 is the mass of steam, Lv1 is the latent heat of condensation of steam, and Qv1 is heat released by 100°C steam of mass m1 to condense into water at 100°C.
The standard value of latent heat of condensation of steam is 540 cal/g. Therefore, substitute 100 g for m1 and 540 cal/g for Lv1.
Qv1=(100 g)(540 cal/g)=54000 cal
Understand that after condensation of all the steam into water at 100°C, water loses heat, which leads to a fall in temperature of the water to attain the final temperature of the mixture.
Write the expression for the change in temperature of condensed steam into water from 100°C to the final temperature of the mixture.
ΔT1=100°C−T1
Here, T1 is the final temperature of the mixture, and ΔT1 is the change in temperature of the steam after condensing into water.
Write the expression for heat lost by water due to change in temperature.
Q1=m1c1ΔT1
Here, Q1 is the heat lost by condensed water and c1 is the specific heat of water.
The standard value of specific heat of water is 1 cal/(g⋅°C). Substitute 100°C−T1 for ΔT1, 1 cal/(g⋅°C) for c1, and 100 g for m1
Q1=(100 g)[1 cal/(g⋅°C)](100°C−T1)=10000−100T1
Calculate the change in temperature of ice and water from 0°C to the final temperature of the mixture.
ΔT2=T1−0°C=T1
Here, ΔT2 is the change in temperature of ice and water from 0°C to the final temperature of the mixture.
Understand that the change in temperature of the calorimeter is same as the change in temperature of water and ice.
ΔTc=ΔT2
Here, ΔTc is the change in temperature of the calorimeter.
Substitute T1 for ΔT2
ΔTc=T1
Write the expression for heat required to convert ice at 0°C into water.
Qf1=miLf1
Here, mi is the mass of ice, Qf1 is the heat required to convert ice at 0°C into water, and Lf1 is the latent heat of fusion of ice at 0°C.
The standard value of latent heat of fusion of ice is 80 cal/g. Substitute 20 g for mi and 80 cal/g for Lf1
Qf1=(20 g)(80 cal/g)=1600 cal
Recall the expression for heat transfer due to temperature difference in order to calculate the heat supplied to change the temperature of water at 0°C formed by the melting of ice to form water at the final temperature of the mixture.
Q3=mic1ΔT2
Here, Q3 is the heat supplied to convert 0°C water to the final temperature of the mixture.
Substitute T1 for ΔT2, 1 cal/(g⋅°C) for c1, and 20 g for mi
Q3=(20 g)[1 cal/(g⋅°C)](T1)=20T1
Recall the expression for heat transfer due to temperature difference in order to calculate the heat removed from 200 g of 0°C water to obtain the final temperature of water in the mixture.
Q2=mwc1ΔT2
Here, mw is the mass of water initially at 0°C and Q2 is the heat removed from the 0°C water to obtain the final temperature of water.
Substitute T1 for ΔT2, 1 cal/(g⋅°C) for c1, and 200 g for mw
Q2=(200 g)[1 cal/(g⋅°C)](T1)=200T1
Understand that the calorimeter is equivalent to 30 g of water. Recall the expression for heat transfer due to temperature difference in order to calculate the heat removed from the calorimeter at 0°C to obtain the final temperature of the mixture.
Q4=m2c1ΔTc
Here, m2 is the mass of the calorimeter and Q4 is the heat removed from the calorimeter to obtain the final temperature of the mixture.
Substitute T1 for ΔTc, 1 cal/(g⋅°C) for c1, and 30 g for m2
Q4=(30 g)[1 cal/(g⋅°C)](T1)=30T1
Recall the expression for net heat transferred in a calorimeter. Take the heat released by the body as positive and the heat absorbed by the body as negative
∑Qnet=0Qv1+Q1−Qf1−Q2−Q3−Q4=0
Substitute 200T1 for Q2, 30T1 for Q4, 20T1 for Q3, 10000−100T1 for Q1, 54000 cal for Qv1, and 1600 cal for Qf1
54000 cal+(10000−100T1)−(1600 cal)−(200T1)−(20T1)−(30T1)=054000+10000−1600=100T1+200T1+20T1+30T162400=350T1T1=62400350
Further solve as,
T1≃178°C
Understand that the final temperature of the mixture cannot be more than 100°C. This indicates that our assumption that the final temperature of the mixture is less than 100°C is incorrect. Therefore, not all the steam condenses. Hence, the final temperature of the mixture is 100°C.
Rewrite the expression for heat released by the mass of steam condensed.
Qv2=msLv1
Here, Qv2 is the heat released by the mass of steam condensed and ms is the mass of steam condensed.
Substitute 540 g for Lv1
Qv2=ms(540 cal/g)=540ms
The final temperature of the mixture is 100°C. This indicates that the heats Q1 and Qv1 will not exist in the final calorimeter heat expression. Rewrite the expression for heat transferred in calorimeter.
∑Qnet=0Qv2−Qf1−Q2−Q3−Q4=0
Substitute 200T1 for Q2, 30T1 for Q4, 20T1 for Q3, 540ms for Qv2, and 1600 cal for Qf1
540ms−(1600 cal)−(200T1)−(20T1)−(30T1)=0540ms−1600=200T1+20T1+30T1540ms=250T1+1600
The final temperature of the mixture is 100°C. Therefore, substitute 100°C for T1.
540ms=250(100°C)+1600540ms=25000+1600540ms=26600ms=26600540
Further solve as,
ms≃49 g
Conclusion:
The result is that 49 g of steam is condensed and the final temperature of the mixture is 100°C.