How much energy is needed to melt lead weighing 20 kg, if its initial temperature is 27 degrees

| February 20, 2021 | education

m = 20 kilograms is the mass of lead;

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

T1 = 327 degrees Celsius – the melting point of lead;

c = 140 J / (kg * C) – specific heat of lead;

q = 25 * 10 ^ 3 J / kg is the specific heat of fusion of lead.

It is required to determine how much energy Q (Joule) is needed to melt lead.

Let’s find the amount of heat that is needed to heat lead to its melting point:

Q1 = c * m * (T1 – T) = 140 * 20 * (327 – 27) = 2800 * 300 = 840,000 Joules.

Let’s find the amount of heat that is needed to melt the lead:

Q2 = q * m = 25000 * 20 = 500000 Joules.

The total amount of heat will be equal to:

Q = Q1 + Q2 = 840,000 + 500,000 = 1,340,000 Joules = 1.34 MJ.

Answer: to melt lead, you need to spend an energy equal to 1.34 MJ.



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