Heat question -- Ice cube in a cup of water....

[zanyzoya]

Homework Statement

An ice cube of mass 0.01.kg at a temperature of 0 degrees Celsius is dropped into a cup containing 0.10kg of water at a temperature of 15 degrees Celsius. What is the max estimated change in temperature of the contents of the cup?
SHC of water = 4200J per kg per kelvin
latent heat of fusion of ice = 340000 J per kg

Homework Equations

and attempt at a solution[/B]

heat lost by water = heat gained by ice
m_waterC ΔT_water = m_ice C ΔT_ice + mΔL

0.1 x 4200 x ΔT_water = 0.01 x 4200 ΔT_ice + 0.01 x 340000

420(T_final-15) = 42 (T_final-0) + 3400
420T - 6300 = 42T + 3400
T= 25 degrees
so change in temperature = 15-25 = -10 degrees celcius
This is wrong as the actual answer is 8.7 K

 

[DrClaude]

T= 25 degrees

So the final temperature is higher than all initial temperatures? Doesn't make sense, right?

heat lost by water = heat gained by ice
m_waterC ΔT_water = m_ice C ΔT_ice + mΔL

First problem: the left-hand side here doesn't correspond to "heat loss by water."

420(T_final-15) = 42 (T_final-0) + 3400

Second problem: you have to use absolute temperatures here.

 

[zanyzoya]

Thanks Dr Claude
Re your first point, my understanding is as follows:
As the water cools down it will loose heat equal to mc delta T, also heat will by lost to surroundings and to the cup. Some of this heat which is lost by the water is given to the ice to initially cause a change in state and finally raise its temperature. Am I on the wrong lines here?

I've converted the values to kelvin and I still get the same answer?!?

 

[DrClaude]

As the water cools down it will loose heat equal to mc delta T, also heat will by lost to surroundings and to the cup. Some of this heat which is lost by the water is given to the ice to initially cause a change in state and finally raise its temperature. Am I on the wrong lines here?

Your approach is correct, but you have to be careful about the signs. (This is a general comment: when you get a final temperature that is higher than the initial temperature, while it should be the other way around, the first thing to check are the signs.)

 

[gneill]

... the actual answer is 8.7 K

That value looks a little high to me. Just melting the ice (ice@0C → water@0C) should take enough heat from the existing liquid water to drop its temperature below that value.

 

[willem2]

That value looks a little high to me. Just melting the ice (ice@0C → water@0C) should take enough heat from the existing liquid water to drop its temperature below that value.

Not that the 8.7K that was asked, is the temperature change and not the final temperature, which is 6.3C.

 

[gneill]

Not that the 8.7K that was asked, is the temperature change and not the final temperature, which is 6.3C.

Ah. My mistake. I should really pay closer attention to these small details. Thanks

 

[haruspex]

heat gained by ice

42 (Tfinal-0)

So far so good: heat gained is proportional to the final temperature minus the initial temperature.
So for heat lost,

heat lost by water

how should you calculate the temperature difference? Not like this:

420(Tfinal-15)