Final answer:
The question involves calculating the change in temperature for water that absorbs heat from an equally massive copper pot cooling down from 275 °C to 25 °C. Using the formula for heat transfer and considering the specific heat capacities of copper and water, we calculate the heat lost by the copper and equal it to the heat gained by the water to find the temperature change of the water.
Step-by-step explanation:
The student is asking about heat transfer and temperature change, which falls under the subject of Physics. This concept is typically taught at the high school level, possibly in a Physics or Chemistry class. The question involves the specific heat capacity of copper and water, and it requires calculating how much the temperature of water would change if it absorbed the heat from a copper pot of equal mass cooling down from 275 °C to 25 °C.
To solve this problem, we use the formula: q = mcΔT, where q is the heat transferred, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature. We know the specific heat capacity of copper is approximately 0.385 J/g°C, and for water it is 4.186 J/g°C. The mass of the copper is given as 536 grams, and its temperature change (ΔT) is 250 °C (275 °C - 25 °C).
The change in temperature for the water, using the heat gained by the water (which is equal to the heat lost by the copper), can be calculated once we know the heat transferred by the copper: q_copper = (536 g)(0.385 J/g°C)(250 °C). We then set this equal to the heat absorbed by the water and solve for the temperature change of the water: q_water = (536 g)(4.186 J/g°C) ΔT.