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(assu...

Assessment: JAMB UTME - Physics - 2025 Subject: Physics

Question 1 Report

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)

Answer Details

Specific heat capacity is the heat needed to raise the temperature of one kilogram of a substance by one kelvin. It comes from the heat equation \[Q = mc\Delta\theta,\] where \(Q\) is the heat supplied in joules, \(m\) the mass in kilograms and \(\Delta\theta\) the temperature rise. Because heat loss to the surroundings is stated to be negligible, all 1000 J supplied goes into the rod, so no correction is needed.

Making \(c\) the subject and substituting: \[c = \frac{Q}{m\Delta\theta} = \frac{1000}{0.025 \times 15} = \frac{1000}{0.375} = 2666.7\ \text{J kg}^{-1}\text{K}^{-1}.\] So the specific heat capacity is \(2666.7\ \text{J kg}^{-1}\text{K}^{-1}\).

Note that the temperature rise needs no conversion. A change of \(15\ ^\circ\text{C}\) is a change of 15 K because the two scales have the same size of degree, so adding 273 here is a wasted step that produces a badly wrong answer. The other frequent slip is working out the denominator carelessly: \(0.025 \times 15 = 0.375\), not 0.0375 or 3.75. Exam reminder: distinguish specific heat capacity \(c\), measured in \(\text{J kg}^{-1}\text{K}^{-1}\), from heat capacity \(C = mc\), measured in \(\text{J K}^{-1}\); the units in the options tell you which one is wanted.

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