A 0.250-kg aluminum bowl holding 0.800 kg of soup at 27.0°C is placed in a freezer. What is the final temperature if 430 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).

-17.3 degres celcius

To find the final temperature of the bowl and soup after energy transfer, we need to consider the heat gained or lost by each component separately.

First, let's calculate the heat lost by the aluminum bowl. We can use the formula:

Q = m * c * ΔT

Where:
Q is the heat transferred
m is the mass of the aluminum bowl
c is the specific heat of aluminum
ΔT is the change in temperature

Given:
m = 0.250 kg
c = 0.215 kcal/(kg · °C)
ΔT (unknown)

We know that 430 kJ is transferred from the bowl, which is equal to 430,000 calories. Converting this value into kcal will give us 430,000 kcal.

Q_aluminum = 430,000 kcal
m_aluminum = 0.250 kg
c_aluminum = 0.215 kcal/(kg · °C)

Rearranging the formula, we can solve for ΔT:

ΔT_aluminum = Q_aluminum / (m_aluminum * c_aluminum)

ΔT_aluminum = 430,000 kcal / (0.250 kg * 0.215 kcal/(kg · °C))

ΔT_aluminum = 8000 °C

Since ΔT_aluminum is a significant value, it means that the aluminum bowl completely solidified and froze during the energy transfer.

Now, let's calculate the heat lost by the soup. We'll consider two parts: the heat lost during cooling and the heat lost during freezing.

The heat lost during cooling:
Q_cooling = m_soup * c_liquid * ΔT_cooling

Given:
m_soup = 0.800 kg
c_liquid = 1.00 kcal/(kg · °C)
ΔT_cooling (unknown)

Q_cooling = Q_total - Q_aluminum

Q_cooling = 430,000 kcal - 430,000 kcal

Q_cooling = 0 kcal

Since the heat lost during cooling is zero, there is no change in temperature for the liquid soup, which means ΔT_cooling is 0°C.

The heat lost during freezing:
Q_freezing = m_soup * L_fusion

Given:
m_soup = 0.800 kg
L_fusion = 79.8 kcal/kg

Q_freezing = m_soup * L_fusion

Q_freezing = 0.800 kg * 79.8 kcal/kg

Q_freezing = 63.8 kcal

Now, let's calculate ΔT_freezing:

ΔT_freezing = Q_freezing / (m_soup * c_frozen)

c_frozen = 0.500 kcal/(kg · °C)

ΔT_freezing = 63.8 kcal / (0.800 kg * 0.500 kcal/(kg · °C))

ΔT_freezing = 159.5 °C

To find the final temperature, we need to sum up ΔT_cooling and ΔT_freezing:

ΔT_final = ΔT_cooling + ΔT_freezing

ΔT_final = 0 °C + 159.5 °C

ΔT_final = 159.5 °C

Therefore, the final temperature of the bowl and soup after the energy transfer is 159.5°C.