A 0.250-kg aluminum bowl holding 0.800 kg of soup at 25.0ºC is
placed in a freezer. What is the final temperature 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?
To answer this question, we need to apply the principles of heat transfer and use the specific heat capacity equation. Here are the steps to find the final temperature:
Step 1: Determine the initial heat content of the bowl and soup.
The initial heat content, Q_initial, is given by Q_initial = mcΔT, where m is the mass, c is the specific heat capacity, and ΔT is the change in temperature. The specific heat capacity of aluminum is 0.897 J/g°C, and we can assume the specific heat capacity of water for the soup (as stated in the question). Note that we need to convert the mass of the bowl to grams.
The mass of the aluminum bowl is 0.250 kg, which is equal to 250 g.
The mass of the soup is 0.800 kg, which is equal to 800 g.
Q_initial = (mcΔT)_bowl + (mcΔT)_soup
Q_initial = (250 g)(0.897 J/g°C)(T_f - 25.0°C) + (800 g)(4.18 J/g°C)(T_f - 25.0°C)
Here, T_f is the final temperature, which we need to solve for.
Step 2: Determine the heat transferred.
The heat transferred, Q_transferred, is given as 377 kJ (kilojoules), which we need to convert to joules.
1 kJ = 1000 J
Q_transferred = 377 kJ = 377,000 J
Step 3: Set up the equation and solve for the final temperature.
We can equate the initial heat content (Q_initial) to the heat transferred (Q_transferred):
Q_initial = Q_transferred
Substituting the values:
(250 g)(0.897 J/g°C)(T_f - 25.0°C) + (800 g)(4.18 J/g°C)(T_f - 25.0°C) = 377,000 J
Simplifying the equation:
(224.25 J/°C)(T_f - 25.0°C) + (3344 J/°C)(T_f - 25.0°C) = 377,000 J
Combining like terms:
2574.25 J/°C(T_f - 25.0°C) = 377,000 J
Dividing both sides by 2574.25 J/°C:
T_f - 25.0°C = 146.625°C
Finally, solving for T_f:
T_f = 146.625°C + 25.0°C
T_f = 171.625°C
So, the final temperature of the bowl and soup is approximately 171.625°C.