Question #87901
A block of lead of mass 100kg in a crucible and at a temperature of 40 degree Celsius was placed in an electric furnace rated 10kw. If the melting point of lead is 320 degree Celsius, calculate the
(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 = 120J/kgk, specific latent heat of fusion of lead = 25000J/kg
1
Expert's answer
2020-04-06T09:31:23-0400

we have


N=0,01(MW)m=100(kg);T1=40(C);T2=320(C);N=0,01 (MW) m=100 (kg); T_1=40 (C); T_2=320 (C);c=120(J/kg);λ=25000(J/kg);Δt=T2T1=280(C)c=120 (J/kg); \lambda=25000 (J/kg); \Delta t=T_2-T_1=280 (C)

1)Quantity of heat required to heat the lead to its melting point

Q1=cmΔtQ_1=cm\Delta tQ1=120100280=3.36(MJ)Q_1=120\cdot100\cdot280=3.36 (MJ)

2)Additional heat energy required to melt the lead

Q2=λmQ_2=\lambda mQ2=25000100=2.5(MJ)Q_2=25000\cdot 100=2.5 (MJ)

3) Time taken to supply this additional energy

t=Q1+Q2Nt=\frac{Q_1+Q_2}{N}t=3.36+2.50.01=586(s)t=\frac{3.36+2.5}{0.01}=586 (s)


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Comments

Assignment Expert
06.04.20, 16:31

Dear Makinde, thank you for your comment

Makinde
06.04.20, 14:39

The solution is incorrect Q2= 25000 x 100 = 2500000J = 2.5MJ and this affect the answer of time t should be t= (2.5+3.36)/0.01 =586s

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