Question

9.(d) A jug of heat capacity 250JK⁻¹ contains water at 28°C. An electric heater of resistance 35Ω connected to a 220V source is used to raise the temperature of the water until it boils at 100°C in 4 minutes. After another 5 minutes, 300g of water has evaporated. Assuming no heat is lost to the surroundings, calculate:
(i) the mass of water in the jug before heating;
(ii) the specific latent heat of vaporization of steam. [Specific heat capacity of water = 4200JK⁻¹kg⁻¹]

Accepted Answer

Heat supplied in first 4min (240s): Q=V²t/R = 220²×240/35 ≈ 331,886J.
This heats the jug (ΔQ=250×72=18,000J) and the water (mcΔT=m×4200×72=302,400m).
331,886 = 18,000 + 302,400m → m ≈ 1.04kg.
(ii) Heat supplied in next 5min (300s): Q₂=220²×300/35 ≈ 414,857J. This evaporates 0.3kg of water:
L = Q₂/mass = 414,857/0.3 ≈ 1.38×10⁶ J/kg.

WAEC Physics 2018 Theory — Question 17

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