What is the amount of heat energy required to convert ice of mass 20 Kg at −40C to water at 200C? Use the relevant option for calculation.
a) Latent heat of fusion of ice 3.34 x 105 J / Kg.
b) Specific heat capacity of water is 4180 J / Kg / K
c) Specific heat of ice 2093 J / Kg / K
Given: ice of mass 20 Kg at −40C to water at 200C
To find: the amount of heat energy required to convert ice to water
Solution:
As per the given condition,
Latent heat of fusion of ice, L=3.34×105J/Kg.
Specific heat capacity of water, s=4180J/Kg/K
Specific heat of ice, c=2093J/Kg/K
Mass of ice, m=20kg
The total heat required will be sum of heat energy needed to change the temperature of ice from –4°C to 0°C plus heat required to undergo a solid to liquid phase change at its melting point plus heat required to heat the sample from 0∘C to 20∘C
qtotal=q1+q2+q3⟹qtotal=mc(ΔT)+mL+ms(ΔT)⟹qtotal=20×2093×(0−(−4))+20×3.34×105+20×4180(20−0)⟹qtotal=167440+66.8×105+1672000⟹qtotal=8519440J=8.52×103kJ
is the quantity of heat is required to transform the ice into water