Answer to Question #87901 in Molecular Physics | Thermodynamics for chuks

Question

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

Answer 1

we have

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

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

$$Q_1=cm\Delta t$$ $$Q_1=120\cdot100\cdot280=3.36 (MJ)$$

2)Additional heat energy required to melt the lead

$$Q_2=\lambda m$$ $$Q_2=25000\cdot 100=2.5 (MJ)$$

3) Time taken to supply this additional energy

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