How much heat is needed to heat 30 g of tin, the temperature of which is 32 C, in order to melt it?

Given:

m = 30 grams is the mass of tin;

t1 = 32 ° Celsius – the initial temperature of the tin;

t2 = 232 ° Celsius – the melting point of tin;

q = 60,000 Joule / kilogram is the specific heat of fusion of tin;

c = 230 J / (kg * C) – specific heat of tin.

It is required to determine Q (Joule) – how much heat must be spent to heat the tin to the melting point and melt it.

Let’s convert mass units from grams to kilograms:

m = 30 grams = 30 * 10-3 = 30/1000 = 0.03 kilograms.

Then the amount of energy to heat the tin will be equal to:

Q1 = c * m * dT = c * m * (t2 – t1) = 230 * 0.03 * (232 – 32) =

= 230 * 0.03 * 200 = 230 * 6 = 1380 Joules.

To melt tin requires energy:

Q2 = q * m = 60,000 * 0.03 = 1800 Joules.

Total energy needed:

Q = Q1 + Q2 = 1380 + 1800 = 3180 Joules.

Answer: Energy required is 3180 Joules (approximately 3200 Joules).