at 1 atm pressure and 0 C temperature 14400 cal of heat is absorbed in fusion of 18g of ice. specific volumes of ice and water is 1.089 and 1.0 ml g^-1 calculate dH and dE
Answers
At 1 atm pressure and 0 C temperature 14400 cal of heat is absorbed in fusion of 18g of ice. Since the pressure is constant , the change in enthalpy is equal to the heat of the reaction.
ΔH =14400 cal
ΔH = ΔE +pΔv
Specific volume of ice= 1.089 ml/g
Volume of ice = specific volume* mass of ice
= 1.089* 18= 19.602ml
Specific volume of water =1ml/g
Volume of water = specific volume* mass of water
=1* 18= 18ml
Δv= 0.0016L
p = 1 atm
pΔv =0.0016Latm= 0.0016*24.217= 0.0387 cal
ΔE = ΔH-pΔv
= 14400- 0.0387=14399.96 cal
Hence the value of ΔE is 14399.96 cal
Explanation:
ΔH =14400 cal
ΔH = ΔE + pΔv
Specific volume of ice= 1.089 ml/g
Volume of ice = specific volume* mass of ice
= 1.089 x 18 = 19.602 ml
Specific volume of water = 1 ml/g
Volume of water = specific volume* mass of water
= 1 x 18 = 18 ml
Δv= 0.0016 L
p = 1 atm
pΔv = 0.0016 L atm = 0.0016 x 24.217= 0.0387 cal
ΔE = ΔH-pΔv
ΔE = 14400- 0.0387 = 14399.96 cal
Hence the value of ΔE is 14399.96 cal