Final answer:
The final temperature of the soup can be calculated by dividing the total energy removed (377,000 J) by the sum of the products of the mass and specific heat capacities of both the soup and the bowl, and then subtracting the result from the initial temperature of the soup.
Step-by-step explanation:
To solve for the final temperature of the soup when 377 kJ of energy is removed, we will use the concept of heat transfer. We assume that the soup's thermal properties are the same as water and use the specific heat capacities of water and aluminum. The specific heat capacity of water is 4.18 J/g°C, and for aluminum, it is 0.897 J/g°C. To simplify calculations, we'll work in joules and convert the 377 kJ of energy removed to 377,000 J.
Let ms be the mass of the soup, cs be the specific heat capacity of the soup, mb be the mass of the bowl, cb be the specific heat capacity of the bowl, ΔT be the change in temperature, and Q be the heat transferred. The equation for heat transfer is Q = (ms × cs + mb × cb) × ΔT.
To find the ΔT, rearrange the equation to ΔT = Q / (ms × cs + mb × cb). Plugging in the values we have: ΔT = 377,000 J / ((800 g × 4.18 J/g°C) + (250 g × 0.897 J/g°C)). After calculating ΔT, we subtract it from the initial temperature of the soup, which is 25.0°C, to get the final temperature. If ΔT is greater than 25.0°C, the final temperature would be below 0°C, and the soup would freeze.