Question

12(c) A block of lead of mass 100 kg in a crucible at a temperature of 40°C was placed in an electric furnace rated 10 kW. If the melting point of lead is 320°C, calculate: (i) quantity of heat required to heat the lead to its melting point (ii) additional heat energy required to melt the lead (iii) time taken to supply this additional energy. (Specific heat capacity of lead = 120 Jkg⁻¹K⁻¹; specific latent heat of fusion of lead = 2.5x10⁴ JK⁻¹)

Accepted Answer

(i) Quantity of heat required to heat the lead to its melting point Q = mLCLΔT = 100x120x(320-40) = 3.36x10^6 J.
(ii) Additional heat required to melt the lead = mLlf where lf = latent heat of fusion of lead. Qf = 100x2.5x10^4 = 2.5x10^6 J.
(iii) Time taken to supply the additional heat: T = additional heat energy introduced / electric power = 2.5x10^6/10x10^3 = 250s.

WAEC Physics 2011 Theory — Question 21

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