JAMB - Physics (2025 - No. 65)
Calculate the specific heat capacity of a metal rod of mass 0.025kg whose temperature was raised by 15ºC when 1000J of heat energy was added to the rod(assuming the heat loss to the surrounding is negligible)
2440.5JKg\(^{-1}\)K\(^{-1}\)
3000.0JKg\(^{-1}\)K\(^{-1}\)
2888.4JKg\(^{-1}\)K\(^{-1}\)
2666.7JKg\(^{-1}\)K\(^{-1}\)
Explanation
D. 2666.7 \(\text{J kg}^{-1} \, \text{K}^{-1}\)
Using the formula: \(Q = mc\Delta T\)
We can rearrange to find \( c \): \(c = \frac{Q}{m \Delta T} = \frac{1000}{0.025 \times 15} = \frac{1000}{0.375} = 2666.7\)
Thus, the specific heat capacity is:
\(c = 2666.7 \, \text{J kg}^{-1} \, \text{K}^{-1}\).
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