An industrial reactor has a heat capacity of 33,400 kJ/°C.How much will the temperature of the reactor rise if it absorbs 3.00 × 10⁵ kJ of heat energy?

Question

Quizplus · Accepted Answer

To calculate the change in temperature of the reactor, we use the formula:
ΔT = Q/C, where Q is the heat energy absorbed and C is the heat capacity of the reactor.
ΔT = (3.00 × 10^15 kJ)/(33,400 kJ/°C) = 8.98°C
Therefore, the temperature of the reactor will rise by 8.98°C. So, the answer is choice C.

An Industrial Reactor Has a Heat Capacity of 33,400 KJ/°C