Multiple Choice
How much heat is needed to change of liquid water at into steam at ?
20. Heat and Temperature
Latent Heat & Phase Changes
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How much heat must be removed from 0.7 kg of water at 23°C to cool it to 0°C and completely freeze it?
A
-5.31 × 106 J
B
1.67 × 105 J
C
- 1.65 × 106 J
D
- 3.01 × 105 J
Verified step by step guidance1
Identify the two processes involved: cooling the water from 23°C to 0°C and then freezing the water at 0°C.
Calculate the heat removed to cool the water from 23°C to 0°C using the formula: Q1 = mcΔT, where m is the mass of the water (0.7 kg), c is the specific heat capacity of water (4,186 J/kg°C), and ΔT is the change in temperature (23°C - 0°C).
Calculate the heat removed to freeze the water at 0°C using the formula: Q2 = mL, where m is the mass of the water (0.7 kg) and L is the latent heat of fusion for water (334,000 J/kg).
Add the two quantities of heat removed: Q_total = Q1 + Q2. This will give the total heat that must be removed to cool and freeze the water.
Ensure the sign of the total heat removed is negative, as heat is being removed from the system, resulting in a negative value for Q_total.
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