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if a 35 kg of ice at -10 degrees C is combined with 110 kg of...

if a 35 kg of ice at -10 degrees C is combined with 110 kg of water at 80 degree C, what will be the final equilibrium temperature (in C) of the system?

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Answer #2

2108*(T+10)*35 = 4181*110*(80-T)

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Answer #3

SOLUTION :


Let tº C be the equilibrium temperature.


Heat absorbed by 35 kg of ice to become water at tº C 

= Heat for temperature riseof ice  from - 10º C to 0º C + latent heat to melt ice

+ heat for raising the temperature of melt water from 0º C to tº C .


= 35 * 0.5 * 10 + 35 * 80 + 35 * 1 * t  CHU 

(sp. heat of ice = 0.5 and that of water =1.0)


= 2825 + 35 t CHU


Heat given up by 110 kg hot water for fall from 80º C to tº C

= 110 * 1 * (80 - t) CHU


Heat absorbed by ice = Heat given up by hot water.

=> 2825 + 35 t = 8800 - 110t 

=> 35t + 110t = 8800 - 2825

=> 145 t = 5975 

=> t = 5975/145 = 41.21º C


So, equilibrium temperature will be = 41.21º C (ANSWER).



answered by: Tulsiram Garg
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