A 0.02 kg ice cube was melted at 0°C by an application of heat which further increased the temperature of the resulted water to 40°C. Calculate the…
Source: JAMB · 2024
A 0.02 kg ice cube was melted at 0°C by an application of heat which further increased the temperature of the resulted water to 40°C. Calculate the total heat supplied within this temperature range. [specific latent heat of fusion of ice = 3.34 × 10⁵ Jkg⁻¹, specific heat capacity of water = 4200 Jkg⁻¹K⁻¹]
Explanation
The heat has two parts: melting the ice at 0°C, then warming the water from 0°C to 40°C.
Melting: Q₁ = mL = 0.02 × 3.34 × 10⁵ = 6680 J.
Heating: Q₂ = mcΔθ = 0.02 × 4200 × 40 = 3360 J.
Total = 6680 + 3360 = 10040 J. Using only the heating part gives 3320 J-type wrong answers, because the latent heat must be included.
Was this explanation helpful?
Something wrong or missing? Tell us — we read every report and fix issues fast.