How much heat is needed to melt a 1 kg piece of lead at an initial temperature of 27 ° C?

We give all the values ​​from given in the SI system:
t1 = 27 ° C = 300 K.
1. The amount of heat required for heating ice and for melting is equal to:
Q = Q1 + Q2, where Q1 is the amount of heat required to heat the body to the melting point, Q2 is the amount of heat required to melt the body.
2. The amount of heat spent on heating the body is equal to the product of the specific heat capacity of the substance, body weight and the difference between the final and initial temperatures.
Q = c * m * (t2-t1), where c is the specific heat capacity of the substance, m is the mass of the substance, t2 and t1 are the final and initial temperatures, respectively.
The specific heat capacity of lead is c = 140 J / kg * K, the melting point of lead is 600 K.
3. To melt the body, the required amount of heat:
Q = λ * m, where λ is the specific heat of fusion of the substance, m is the mass of the substance.
Specific heat of fusion of lead λ = 0.25 * 10 ^ 5 J / kg.
4. Substitute everything into formula 1:
Q = Q1 + Q2 = c * m * (t2-t1) + λ * m.
Substitute the numerical values:
Q = c * m * (t2-t1) + λ * m = 140 * 1 * (600-300) + 0.25 * 10 ^ 5 * 1 = 0.67 * 10 ^ 5 J.
Answer: 0.67 * 10 ^ 5 J or 67 kJ.