A piece of copper of mass 7.5 g at `27^(@)C` is dropped in boling liquid oxygen (boling point `-183^(@)C`). The released oxygen oc cupies 1.89 letres at `20^(@)C` and a pressure of 750 mm. Find the latent heat of vaporisation of oxygen. Given that specific heat of copper=`0.08 cal g^(-1) .^(@)C^(-1)` and density of oxygen at NTP is `1.429 g//litre`.

Explanation

Here, mass of piece of copper, `M=7.5 g`
Fall in temperature , `Delta T = 27 -(-183)=210^(@)C`
Sp. Heat of copper, `s=0.08 cal g^(-1 .^(@)C^(-1)`
Heat lost by copper, `Q = ms Delta T = 7.5 xx 0.08 xx 210`
`=126 cal`.
Let on dropping the piece of copperinto liquid oxygen, m gram of oxygen evaporates. if L is the latent heat of vaporisation of liquid oxygen, then heat gained by liquid oxygen
`Q' = mL`
From principle of calorimetry ltbr. `Q = Q' or 126 = mL` ..(i)
The mass of liquid oxygen evaporated can be determined by considering the volume of the oxygen released at `20^(@)C` to its volume at N.T.P. Let 1.89 litres of oxygen at `20^(@)C` and 750 mm pressure oc cupies a volume `V_(0)` in litres at N.T.P.
As `(P_(0)V_(0))/(T_0)` or `(760 xx V_(0))/(273) = (750 xx 1.89)/((273+20))`
or `V_(0) = (750 xx 1.89 xx 273)/(293 xx 760) = 1.738 litres`.
since density of oxygen at NTP is `1.429 g litre^(-1)` so mass, `m=V_(0)rho = 1.738 xx 1.429 = 2.484 g`
From equation (i)
`L =(126)/(m) = 126/(2.484) = 50.725 cal g^(-1)`.