NEET-XII-Physics
pp year:2018
- #9A sample of 0.1 g of water at 100°C and
normal pressure (1.013 x ``10^5`` ``Nm^{-2}``)
requires 54 cal of heat energy to convert
to steam at ``100^o``C. If the volume of the
steam produced is 167.1 cc, the change in
internal energy of the sample, is :-
(1) 104.3 J
(2) 208.7 J
(3) 42.2 J
(4) 84.5 JdigAnsr: 2Ans : (2)
Sol. ▵Q = 54 cal = 54 × 4.18 joule = 225.72 joule
▵W = P[Vsteam - Vwater] [For water 0.1 gram=0.1 cc]
= 1.013 × 105[167.1 × 10-6 - 0.1 × 10-6] joule
= 1.013 × 167 × 10-1 = 16.917 joule
By FLOT
&implies; ▵U = ▵Q - ▵W = 225.72 - 16.917
▵U = 208.8 joule