Answer:
First, we need to calculate the amount of heat transferred when the frozen piece of alcohol melts and then heats up to the final temperature.
The heat transferred, Q, can be calculated using the formula:
Q = mcΔT
where:
m = mass of the frozen alcohol (10 g)
c = specific heat of the alcohol (0.57 cal/g°C)
ΔT = change in temperature (7°C - (-11°C) = 18°C)
Q = (10 g)(0.57 cal/g°C)(18°C) = 102.6 cal
Next, we need to calculate the amount of heat released by the liquid alcohol as it cools from 22°C to 7°C.
The heat released, Q, can be calculated using the formula:
Q = mcΔT
where:
m = mass of the liquid alcohol (150 g)
c = specific heat of the alcohol (0.57 cal/g°C)
ΔT = change in temperature (22°C - 7°C = 15°C)
Q = (150 g)(0.57 cal/g°C)(15°C) = 128.25 cal
Since the heat released by the liquid alcohol (128.25 cal) is greater than the heat absorbed by the frozen alcohol (102.6 cal), we can assume that the excess heat released by the liquid alcohol is due to the latent heat of fusion of the frozen alcohol.
The latent heat of fusion (L) can be calculated using the formula:
L = Q/m
where:
Q = excess heat released by the liquid alcohol (128.25 cal - 102.6 cal = 25.65 cal)
m = mass of the frozen alcohol (10 g)
L = 25.65 cal / 10 g = 2.565 cal/g
Therefore, the latent heat of fusion for this alcohol is 2.565 cal/g.