JAMB - Physics (2025 - No. 46)

Calculate the heat capacity of a material that absorbs 48KJ of heat at a differential temperature of 53ºC

760.8JK\(^{-1}\)

2500JK\(^{-1}\)

905.7JK\(^{-1}\)

260.5JK\(^{-1}\)

\(C = \frac{Q}{\Delta T} = \frac{48000 \, \text{J}}{53 \, \text{ºC}} \approx 905.66 \, \text{J/ºC}\)