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navm1
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Homework Statement
If a steel ball of mass 100g and 100°c is dropped into 1litre of water at 20°c, what is the temperature rise? What would happen if it were dropped into a mixture of ice and water at 0°c?
Homework Equations
Q=mcΔT
Q=mlm where ml is latent heat of melting
Q lost = Q gained
The Attempt at a Solution
Qsteel = (0.1) (420) (-80) = -3360J
the steel ball loses 3360J so the water must have gained 3360J
Qwater = 3360 = (1)(4200)ΔT
ΔT = 3360/4200=0.8°C
so the waters temperature has increased by 0.8 degrees.
For the second part it didnt specify how much ice so i just chose to have 1 litre of water with 150g of ice at 0°c.
Qsteel = (0.1)(420)(-100) = -4200J
so the steel ball has lost 4200J of heat and the water ice mixture will gain 4200J of heat
Qwater = Qice = mlm
mice= Qwater/lm = 4200/3.35x105 = 0.0125kg
So only 12.5g of ice would be melted and the temperature wouldn't rise at all as the energy has gone into melting the ice and there is still a significant amount left.Am I on the right track here? Thanks