Question

If 4520 kJ of heat is needed to boil a sample of water, what is the mass of the sample of water

Answer 1

q = 4520 kJ = 4520000 J.

Solution:

For water at its normal boiling point of 100 ºC, the heat of vaporization (H_v) is 2260 J g^-1. This means that to convert 1 g of water at 100 ºC to 1 g of steam at 100 ºC, 2260 J of heat must be absorbed by the water.

Mathematically, this is expressed as follows:

q= m * Hv;

Therefore,

m(H₂O) = q / H_v = (4520000 J ) / (2260 J/g) = 2000 g = 2 kg.

m(H₂O) = 2000 g = 2 kg.

Answer: 2 kg is the mass of the sample of water.