Question

Find the mass of water that vaporizes when 3.78 kg of mercury at 248 °C is added to 0.260 kg of water at 97.2 °C.

Answer 1

Solution :-

The heat energy is initially required to heat the water to its boiling point.

Q=ms(T₂-T₁)

Mercury can supply heat= 3.78140(248-100)= 78.32 kJ.

The water needs the heat= 0.260 4.184K(100-97.2)= 3.046

So, 78.32- 3.046= 75.274 kJ heat is available for vaporisation.

Mass of water that vaporises= 75.274/2260 where 2260kJ/kg is the latent heat of water.

Mass= 0.0333 kg.