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

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