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\(^{-1}\)
Explanation
\(C = \frac{Q}{\Delta T} = \frac{48000 \, \text{J}}{53 \, \text{ºC}} \approx 905.66 \, \text{J/ºC}\)
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