Calculate the heat capacity of a material that absorbs 48KJ of heat at a differential temperature of 53ºC
A
760.8JK\(^{-1}\)
B
2500JK\(^{-1}\)
C
905.7JK\(^{-1}\)
D
260.5JK\(^{-1}\)
Verified Correct Answer: C.
Physical Principles & Numerical Solution
\(C = \frac{Q}{\Delta T} = \frac{48000 \, \text{J}}{53 \, \text{ºC}} \approx 905.66 \, \text{J/ºC}\)
Source: Myschool Past Questions Archive (Record #74480)