Question

# A 0.23-kg aluminum bowl holding 0.75 kg of soup at 25.0°C is placed in a freezer....

A 0.23-kg aluminum bowl holding 0.75 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?

Suppose the final temperature is T.

Energy required to change the temperature of system to 0 deg C,

Q = m2*Lf + m1*C1*dT1 + m2*C2*dT2

here, dT1 = 25.0

dT2 = 25.0

m1 = mass of aluminium = 0.23 kg

m2 = mass of soup = 0.75 kg

C1 = specific heat of aluminium = 900

C2 = specific heat of soup = 4186

Lf = Latent heat = 3.33*10^5

So, Q = 0.23*900*25 + 0.75*4186*25.0 + 0.75*3.33*10^5

Q = 333.412 kJ

after freeze the soup, we have (377 - 333.412) kJ = 43.588 kJ heat.

Now determine the final temperature of frozen soup using energy conservation:

Q1 + Q2 = 43.588 kJ

m1*C1*dT1 + m2*C2*dT2 = -43588

dT1 = Tf - Ti = T - 0

dT2 = T - 0

Now using given values:

0.23*900*(T -0) + 0.75*2108*(T - 0) = -43588

Now Solving above equation

T = (-43588)/(0.23*900 + 0.75*4186)

T = -13.02 deg C

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