Question

(8%) Problem 3: A 0.275-kg aluminum bowl holding 0.65 kg of soup at 25.0°C is placed in a freezer.What is the final temperature, in degrees Celsius, if 377 kJ of energy is transferred from the bowl and soup, assuming the soup’s thermal properties are the same as that of water? Specifically, you can assume the latent heat of fusion for the soup is 334 kJ/kg.

Answer 1

Total amount of Energy transferred from the bowl and Soup Qnet = 377 kJ

Now first temperature of bowl and soup will decrease to 0 C,

So to bring the system to 0 C, required heat will be

Q = Q1 + Q2

Q1 = Energy required for Soup from 25 C to 0 C = Ms*Cs*dT

Q2 = Energy required for bowl from 25 C to 0 C = Mb*Cb*dT

dT = 0 - 25 = -25

Ms = 0.65 kg & Mb = 0.275 kg

Cs = 4186 J/kg-C & Cb = 900 J/kg-C

So,

Q = 0.65*4186*(-25) + 0.275*900*(-25) = -74210 J = -74.21 kJ

Now remaining energy = Qnet - Q = 377 - 74.21 = 302.79 kJ

Which means Soup will freeze, So now energy required for soup to freeze at 0 C, will be

Q3 = Ms*Lf

Lf = latent heat of fusion for Soup = 334 kJ/kg

Q3 = 0.65*334 = 217.1 kJ

Now remaining energy = 302.79 - 217.1 = 85.69 kJ

So temperature will further go down,

Now Suppose final temperature is = T, and we have total remaining energy 85.69 kJ

Q = Q4 + Q5

Q4 = energy required for soup to reduce temperature from 0 C to T = Ms*Ci*dT

Q5 = energy required for bowl to reduce temperature from 0 C to T = Mb*Cb*dT

Ci = Specific heat of ice (freezed Soup) = 2090 J/kg-C

dT = 0 - T = -T

So,

Q = 85.69 kJ = 85690 J

85690 = 0.65*2090*(-T) + 0.275*900*(-T)

T = -85690/(0.65*2090 + 0.275*900)

T = -53.36 C = FInal temperature of Soup

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