`A` vessel of water eqivalent `W` kg contains m kg of water of specific heat `S` . When water evaporates at the vessel and water of `alpha kgS^(-1)` , the temprature of the vessel and water and in it falls from `T_(1)^(@)C` to `T_(2)^(@)` in t s. If `m gt gt alpha` t and a fracrtion `E` of the heat vessel and the water then average rate of fall of temprature is
(`L=` average latent heat of vapourisation in `J kg^(-1)`)

Options

A. `(in alpha L)/(ms+w).^(@)C//S`
B. `(in alpha )/(L(ms+w)).^(@)C//s`
C. `(in alpha L)/(w).^(@)C//s`
D. `(in alpha L)/(ms).^(@)C//s`

Answer

A

Explanation

Where `L` is the average latent heat of vaporisation in `J kg^(-1)`
As water evaporates at a rate a kg `s^(-1)` , in t ceconds the mass of water evaporation is m' `=alphat kg` . If`L` is the average latent heat of vaporization over the temperature range `T_(1)` to `T_(2)` , then heat required for the evaporation of m' kg of water is `Q=m' L=alpha tL`
Since a fraction `varepsilon` of this heat is taken from the vessel and the water in it, the heat lost from the vessel+water is
`Q'=varepsilon alpha tL`.......(1)
When this heat is lost, the temperature of the vessel and water falls from `T_(1)` to `T_(2)`. Now, at time t, the mass of water lift in the vessel is `(m-alphat)` kg Therefore, heat lost by water and vessel is
`Q"=[(m-alphat)s+W](T_(1)-T_(2))`
Since `alpha t lt lt m,(m-alphat)~=m` . Thus `Q"=(ms+w)(T_(1)-T_(2))` ......(2)
Eqiating Equations `(1) and (2)` we have `(ms+w)(T_(1)-T_(2))=varepsilon proptL`
Rate of fall of temperature is `1(t_(1)T_(2))/(t)=(varepsilon prop L)/((ms+w))`