To find the quantity of heat is required to transform 2kg of water at 20∘C into vapour at 100∘C.
Solution:
We know,
Specific heat of water, s=4.18J/(g∘C)
Enthalpy of vapourization, L=2.3×103kJ/kg=2300J/g
As per the given condition,
Mass of water, m=2kg=2×103g
The total heat required will be sum of heat required to to warm the water from 20.0°C to 100.0°C plus heat required to vapourize the water to steam at 100°C
qtotal=q1+q2⟹qtotal=ms(ΔT)+mL⟹qtotal=2×103×4.18(100−20)+2×103×2300⟹qtotal=668.8×103+4600×103⟹qtotal=5268.6×103J=5269kJ
is the quantity of heat is required to transform 2kg of water at 20∘C into vapour at 100∘C.