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i)Derive an expression for the heat lost by the steam as it condenses to water at temperature T. ii)Derive an expression for the heat gain in...

Steam of mass 3.0g at 100 degress celsius is passed into water of mass 400g at 10 degrees celsius. the final temperature of the mixture is T. The container absorbs negligible heat. (specific latent heat of vaporization of steam = 2260KJ/kg, specific heat capacity of water =4200J Kg-1 K-)

Answers


Jacktone
(i)Heat lost=mLv+mc(100-T)
H=(0.003×2260000)+
(0.003×4200[100-T])
H=6780+1260-12.6T

(ii)Heat gained=Mc(T-10)
H=0.4×4200(T-10)
H=1680T-16800
(iii) 1680T-16800=6780+1260-12.6T
1692.6T=24840
T= 14.68°C
jacktonenzoya answered the question on August 29, 2019 at 06:53

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