Question

A 0.250-kg aluminum bowl holding 0.800 kg of soup at 27.6°C is placed in a freezer. What is the final temperature if 424 kJ of energy is transferred from the bowl and soup? Assume the soup has the same thermal properties as that of water, the specific heat of the liquid soup is 1.00 kcal/(kg · °C), frozen soup is 0.500 kcal/(kg · °C), and the latent heat of fusion is 79.8 kcal/kg. The specific heat of aluminum is 0.215 kcal/(kg · °C).

Answer 1

Answer:

16.32 °C

Explanation:

We are given;

Mass of aluminum bowl; m_b = 0.25 kg

Mass of soup; m_s = 0.8 kg

Thus, formula to find the amount of heat energy for a temperature change of 27.6°C to 0°C is;

Q = (m_b•c_b•Δt) + (m_s•c_s•Δt)

Where;

c_b = 0.215 kcal/(kg•°C)

c_s = 1 kcal/(kg•°C)

ΔT = 27.6 - 0 = 27.6°C

Thus;

Q = (0.25 × 0.215 × 27.6) + (0.8 × 1 × 27.6)

Q = 23.5635 Kcal

Now, the energy that exits to be used to freeze the soup is;

Q' = 424 kJ - Q

Let's convert 424 KJ to Kcal

424 KJ = 424/4.184 Kcal = 101.3384 Kcal

Thus;

Q' = 101.3384 - 23.5635

Q' = 77.7749 Kcal

Amount of heat that's removed is given by;

Q_f = Q' - mL

Where;

m = m_s = 0.8 kg

L = 79.8 kcal/kg

Thus;

Q_f = 77.7749 - (0.8 × 79.8)

Q_f = 13.9349 Kcal

Then final temperature will be;

T_f = Q_f/((m_b•c_b) + (m_s•c_s))

T_f = 13.9349/((0.25 × 0.215) + (0.8 × 1))

T_f = 16.32 °C