A solid sample is heated by an electrical heater of constant power. The temperature-time graph of the process shows that the temperature of the solid rises linearly from 300 K to its melting point of 500 K in 2 minutes . It then remains constant at 500 K for the next 5 minutes until the sample melts completely. If the specific heat capacity of the solid is 400 J kg ⁻¹ K ⁻¹ , the latent heat of fusion of the material

Step-by-step solution

Let the constant power of the heater be P and the mass of the sample be m . During the first 2 minutes ( t₁ ), the heat supplied raises the temperature of the solid: P t₁ = m c T P m = c (T_ melt - T_ initial ) t₁ During the next 5 minutes ( t₂ ), the heat supplied melts the solid at constant temperature: P t₂ = m L P m = L t₂ Equating the two expressions for P m : L t₂ = c (T_ melt - T_ initial ) t₁ Rearranging for the latent heat of fusion L : L = c (T_ melt - T_ initial ) t₂ t₁ Substitute the given values ( c =