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?
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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