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Calculate the specific heat capacity of a metal rod of mass 0.025kg whose temperature was...

Physics
JAMB 2025

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)

  • A. 2440.5JKg\(^{-1}\)K\(^{-1}\)
  • B. 3000.0JKg\(^{-1}\)K\(^{-1}\)
  • C. 2888.4JKg\(^{-1}\)K\(^{-1}\)
  • D. 2666.7JKg\(^{-1}\)K\(^{-1}\)
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Correct Answer: Option D
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:
\(c = 2666.7 \, \text{J kg}^{-1} \, \text{K}^{-1}\).

There is an explanation video available below.


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Explanation Video

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Post-UTME Past Questions - Original materials are available here - Download PDF for your school of choice + 1 year SMS alerts
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