At `0^(@)C` and normal atmospheric pressure, the volume of 1 gram of water increases from `1 c.c "to" 1.091 c.c` on freezing. What will be the change in its internal energy? Normal atmospheric pressure is `1.013xx10^(5)N//m^(2)` and latent heat of melting of ice `= 80 cal//gram`.

Answer

`-80.0022 cal`

Explanation

Here, `dV= 1.091-1= 0.091 c c`
`=0.091xx10^(6)m^(3)`
`P=1 atm = 1.013xx10^(5)N//m^(2) : L = 80 cal//g`
Heat given out by `1 g` of water on frezzing,
`dQ= -mL = -1xx80 cal`
External work done by water in freezing
`dW= pdV= 1.013xx10^(5)xx0.091xx10^(-6)`
`=0.0092= (0.0092)/(4.18)cal. = 0.0022 cal`.
`dU= dQ-dW= -80-0.0022 cal`.
`dU= dQ-dW= -80-0.0022= -80.0022 cal`.