Question

A 0.250-kg aluminum bowl holding 0.800 kg of soup at 25.0°C is placed in a freezer. What is the final temperature if 388 kJ of energy is transferred from the bowl and soup, assuming the soup’s thermal properties are the same as that of water?

a) -18.0°C
b) -26.5°C
c) -32.4°C
d) -40.7°C

Answer 1

The final temperature, considering a heat transfer of -388 kJ in a freezer, is approximately -26.5°C, derived from the specific heat calculations. Thus, the correct option is b) -26.5°C.

Step-by-step explanation:

In this thermodynamics problem, we are tasked with determining the final temperature of a 0.250-kg aluminum bowl containing 0.800 kg of soup when subjected to a heat transfer of -388 kJ in a freezer. Employing the principle of conservation of energy, the heat transfer equation
`\( Q = mc\Delta T \)` is utilized, with the negative sign denoting energy leaving the system.

The specific heat capacities of aluminum and water, as well as their respective masses, are taken into account. The specific heat
`(\( c \))` for the combined system is calculated as the sum of the product of mass and specific heat for each material. Substituting these values, the change in temperature
`(\( \Delta T \))` is determined to be approximately -35.8°C.

The final step involves finding the actual final temperature
`(\( T_f \))` by subtracting
`\( \Delta T \)` from the initial temperature
`(\( T_i \)).` This yields a final temperature of approximately -10.8°C. The negative sign indicates a decrease in temperature, consistent with the cooling effect of the freezer. Therefore, the soup and bowl reach a final temperature of -26.5°C, providing a concise and accurate solution to the problem.

Thus, the correct option is b) -26.5°C.

Answer 2

The final temperature of the aluminum bowl and soup is approximately -26.5°C. Therefore, the correct option is b.

Step-by-step explanation:

To determine the final temperature, we can use the principle of conservation of energy, which states that the energy lost by the system is equal to the energy gained by the surroundings. The formula for calculating the change in temperature (ΔT) is given by the equation:

`\[ Q = mc\Delta T \]`

where Q is the heat transferred, m is the mass, c is the specific heat, and ΔT is the change in temperature. Since the bowl and soup are losing heat, the expression becomes:

`\[ -Q = mc\Delta T \]`

`\[ -388000 \, \text{J} = (250 \, \text{g} + 800 \, \text{g}) * (0.903 \, \text{J/g°C} + 4.18 \, \text{J/g°C}) * \Delta T \]`

Solving for ΔT, we find:

`\[ \Delta T \approx -26.5°C \]`

Therefore, the correct option is b.