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

Assessment: JAMB UTME - Physics - 2025 Subject: Physics

Question 1 Report

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

Answer Details

Heat capacity (thermal capacity) is defined as the heat energy needed to raise the temperature of a whole body by one kelvin:

\[C = \frac{Q}{\Delta\theta}.\]

Its unit, \(\text{J K}^{-1}\), is itself a reminder of the definition: joules per kelvin of temperature change.

Convert the energy to joules before dividing, since the answer is wanted in \(\text{J K}^{-1}\):

\[Q = 48\ \text{kJ} = 48\,000\ \text{J}.\]

A temperature difference of \(53\ ^\circ\text{C}\) is the same size as a difference of \(53\ \text{K}\), because one Celsius degree and one kelvin represent the same interval; only the zero points of the two scales differ. There is therefore no need to add \(273\) to a temperature change. Substituting,

\[C = \frac{48\,000}{53} = 905.7\ \text{J K}^{-1}.\]

Two errors are worth guarding against. Adding \(273\) to the \(53\) gives \(326\ \text{K}\) and a much smaller capacity, and leaving the energy as \(48\ \text{kJ}\) gives \(0.906\), which is in \(\text{kJ K}^{-1}\) rather than the requested unit. Note also that this quantity applies to this particular body only; to obtain the specific heat capacity of the material you would additionally divide by the mass, using \(c = C/m\).

Download The App On Google Playstore

Everything you need to excel in your exams

Green Bridge CBT Mobile App
Personalized AI Learning Chat Assistant
200,000+ Exam Questions Across IGCSE, JAMB, WAEC & NECO
Over 3,900 Lesson Notes
Offline Support - Learn Anytime, Anywhere
Green Bridge Timetable
Literature Summaries & Potential Questions
Track Your Performance & Progress
In-depth Explanations for Comprehensive Learning