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.