`1.0 m^(3)` of water is converted into `1671 m^(3)` of steam at atmospheric pressure and `100^(@)C` temperature. The latent heat of vaporisation of water is `2.3xx10^(6) J kg^(-1)`. If 2.0 kg of water be converted into steam at atmospheric pressure and `100^(@)C` temperature, then how much will be the increases in its internal energy? Density of water `1.0xx10^(3) kg m^(-3)`, atmospheric pressure =`1.01xx10^(5) Nm^(-2)`.

Answer

`4.26`

Explanation

Heat gained by 2.0 kg of water in changing into steam,
`Q = mL = 2.0kg xx 2.3xx10^6 "J kg"^(-1)= 4.6xx10^6 ` J
Volume of 2.0kg water
`=("Mass")/("Density ") = (2.0 kg)/(10^3 "kg m"^(-3)) =2.0xx10^(-3) m^3`
Volume of steam formed by `1.0m^3` of water `= 1671m^3`
`:.` Volume of steam formed by `2.0 xx10^(-3)m^3` of water ` =1671xx2.0xx10^(-3) = 3342xx10^(-3)m^3`
When 2.0 kg of water changes into steam, the increase in volume is dV `= 3342xx10^(-3)-2.0xx10^(-3) = 3340 xx10^(-3)m^3.`
The external work done against the atmospheric pressure, dW = PdV `=1.01xx10^5 xx 3340 xx 10^(-3)`
` = 0.337xx10^6 J`
From first law of thermodynamics, the increase in internal energy is `dU = Q-dW = 4.6xx10^6 – 0.337xx10^6`
` = 4.263xx10^6 ~~ 4.26xx10^6J`
The positive value of dU indicates that the internal energy of ice increases on melting.