Question

Based on the thermodynamic properties provided for water, determine the energy change when the temperature of 0.650 kg of water decreased from 101 °c to 51.0 °c.

Answer 1

The boiling point of water is 100°C. So at 101°C, the water is steam. Compute the specific heat first from 101 to 100.

E = mCΔT, where c for steam is 1.996 kJ/kg·°C
E₁ = (0.65 kg)(1.996 kJ/kg·°C)(101 - 100°C) = 1.2974 kJ

Next, let's solve the latent heat when steam turns to liquid. The heat of vaporization of water is 2260 kJ/kg.

E₂ = mHvap = (0.65 kg)(2260 kJ/kg) = 1469 kJ

Lastly, let's solve the energy to bring down the temperature to 51°C. The specific heat of liquid water is 4.187 kJ/kg·°C.
E₃ = (0.65 kg)(4.187 kJ/kg·°C)(100 - 51°C) = 139.36 kJ

Thus,
Total energy = 1.2974 kJ+1469 kJ+139.36 kJ = 1,609.66 kJ