How much heat is needed to melt 10 kg of lead at a temperature of 27 degrees.

Given:

m = 10 kilograms is the mass of lead;

T1 = 27 degrees Celsius – the initial temperature of the lead;

T2 = 327 degrees Celsius – the melting point of lead;

c = 130 J / (kg * C) – specific heat of lead;

q = 25000 J / kg is the specific heat of fusion of lead.

It is required to determine Q (Joule) – the amount of heat required to melt lead.

Since lead is at 27 degrees Celsius, it must first be heated to its melting point and then melted:

Q = Qheating + Qmelting;

Q = c * m * (T2 – T1) + q * m;

Q = 130 * 10 * (327 – 27) + 25000 * 10 = 1300 * 300 + 250000 =

= 390,000 + 250,000 = 640,000 Joules = 640 kJ.

Answer: to melt lead, you need to expend energy equal to 640 kJ.