Three 102.0-g ice cubes initially at 0°C are added to 0.810 kg of water initially at 18.0°C in an insulated container.
(a) What is the equilibrium temperature of the system?
(b) What is the mass of unmelted ice, if any, when the system is at equilibrium?
heat lost by water in melting ice to 0deg water Qloss = Mw*Sw*(18-0) = = 61031.88 J
heat required to convert ice at 0 to water at 0 Qreq =
Mice*L = 3*0.102*334*10^3 = 102204 J
all the ice melts will not melt
(a)
some ice is unmelted
the equilibrim temperatre is 0oC
<<<+=====ANSWER
(b)
let m kg of ice melts
In equilibrium
heat loss by water = heat gained by m kg of
ice
heat loss by water = Mw*Sw*(18)
m*L = Mw*Sw*18
m*334*10^3 = 0.81*4186*18
m = 0.183 kg
mass of ice unmelted = (3*0.102) - 0.183 = 0.123
kg <<<+=====ANSWER
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