Final answer:
Using the formula Q = mcΔT, we find that pouring 15 grams of 100°C liquid water that cools down to 37°C on a hand absorbs 945 calories of heat. The provided answer choices do not match the correct calculation.
Step-by-step explanation:
The question asks for the amount of heat absorbed when pouring 15 grams of 100°C liquid water on a student's hand, which is subsequently cooled to body temperature (37°C). To calculate the heat transfer, we use the formula:
Q = mcΔT
where Q is the heat absorbed, m is the mass of the water, c is the specific heat capacity of water, and ΔT is the change in temperature. The specific heat capacity (c) for water is 1.00 cal/g°C. The mass (m) is 15 grams, and the temperature change (ΔT) is 100°C - 37°C = 63°C.
Therefore, the calculation is as follows:
Q = (15 g)(1.00 cal/g°C)(63°C)
Q = 945 calories
However, none of the answer choices provided (a) 63.75 calories, (b) 93.75 calories, (c) 110.25 calories, or (d) 132.25 calories match the calculated result of 945 calories. This suggests there might be an error in the question or the answer choices. The correct approach to determining the heat absorbed has been demonstrated.