Heat is supplied uniformly at the rate of 100W to 1.0 x 10 ^ -2 kg of a liquid for 20 seconds. If the…

Analyze the question Heat is supplied at 100 W for 20 s to 1.0 × 10⁻² kg of liquid, and the temperature rises by 5 °C. We need the specific heat capacity, c.

Concept Energy supplied: Q = P × t Heat absorbed: Q = mcΔθ So c = Pt / (mΔθ)

The unit of specific heat capacity is J kg⁻¹ K⁻¹.

Solution

1. Q = 100 × 20 = 2000 J
2. c = Q / (mΔθ) = 2000 / (1.0 × 10⁻² × 5)
3. c = 2000 / (5 × 10⁻²) = 4.0 × 10⁴ J kg⁻¹ K⁻¹

Checking the options

• A and B give 2.0 × 10², which is wrong. It would come from forgetting to divide by the mass properly.
• D has the right number but the wrong unit. It is missing K⁻¹, so it is not a specific heat capacity.
• C gives 4.0 × 10⁴ J kg⁻¹ K⁻¹, which matches both the value and the unit.

Tip Always check the units as well as the number. JAMB often includes options with the right value but a wrong unit, like D here. Remember the chain: Power × time = energy = mcΔθ.