# A piece of lead weighing 2 kg at a temperature of 27 ° C was heated to the melting temperature

**A piece of lead weighing 2 kg at a temperature of 27 ° C was heated to the melting temperature, and 0.5 kg of lead was melted. What amount of heat was required?**

First of all, we will briefly write down the condition given.

So, given:

m = 2 kg (piece of lead);

t1 = 27 ° C;

t2 = 327 ° C (melting point of lead).

c = 140 J / kg * ° С.

lambda = 0.25 * 10 ^ 5 J / kg.

it is also known that 0.5 kg of lead has melted.

To calculate the amount of heat, we will apply the formula:

Q = Q1 + Q2;

First of all, we must heat a piece of lead to its melting point:

Q1 = cm (t2 – t1).

Substitute the values and calculate:

Q1 = 140 * 2 * (327 – 27) = 84,000 J.

then 0.5 kg of lead is melted:

Q2 = lambda * m = 0.25 * 10 ^ 5 * 0.5 = 12 500 J.

Total heat:

Q = Q1 + Q2 = 84,000 + 12,500 = 96,500 J = 96.5 kJ.