we have
N=0,01(MW)m=100(kg);T1=40(C);T2=320(C);c=120(J/kg);λ=25000(J/kg);Δt=T2−T1=280(C) 1)Quantity of heat required to heat the lead to its melting point
Q1=cmΔtQ1=120⋅100⋅280=3.36(MJ) 2)Additional heat energy required to melt the lead
Q2=λmQ2=25000⋅100=2.5(MJ) 3) Time taken to supply this additional energy
t=NQ1+Q2t=0.013.36+2.5=586(s)
Comments
Dear Makinde, thank you for your comment
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