A gas expands from `3 dm^(3)` to `5 dm^(3)` against a constant pressure of 3 atm. The work done during expansion is used to heat 10 mole of water at temperature 290 K. Calculate final temperature of water. Specific heat of water = `4.184 J//g//K`.
(given : 1 atm = 101.33 J)

Explanation

Work done for an irreversible expansion at constant pressure
`W = - P(V_(2)-V_(1))`
`W = - 3xx(5-3)=-3xx2` litre atm `=-6xx101.3` joule = - 607.8 joule
Noe this work is used up in heating water
`therefore W=q_(p)=607.8` joule
`W=C_(p)xxm(dT)`
`607.8 = 4.184xx10xx18xxdT`
`dT=0.80`
`T_(2)-T_(1)=0.80`
`T_(2)=290+0.80=290.80 K`