The change of energy on freezing `1.00 kg` of liquid water at `0^(@)C` and1 atmis

Options

A. `236.7kJ kg^(-1)`
B. `333.4kJ kg^(-1)`
C. `-333.4kJ kg^(-1)`
D. `-236.7kJ kg^(-1)`

Answer

C

Explanation

`W=1kg, T_(0)=0^(@)C=273K, K_(f)=1.86K-kg//"mole"`
`l_(f)` = latent heat of fusion in cal/g.
`l_(f)=(0.002T_(0)^(2))/(k_(f))`
`=(0.002xx273xx273)/(1.86)"cal/g"`.
= 80 cal / g
`=80xx(4.18)/(10^(3))kJxx(1)/(10^(-3))kg^(-1)`
`=334.4 kJ kg^(-1)`
So `l_(f)` amount of energy is required to freeze `0^(@)C` of water to `0^(@)C` ice. So change of energy of liquid water `=-334.4kJ kg^(-1)`.