# Ice cube in hot water

We have a glass containing 0.5 liter (0.5 kg) of water whose temperature 100 degrees Celsius.
We also have an ice cube with mass 0.01 kg and temperature -10 degrees Celsius.

The cube is put into the glass. The glass is then insulated from the outside world, until the cube has melted. What will be the terminal temperature of the water?

I have read a little bit about Thermodynamics and I think this solution shoudl work:

Let's first calculate the amount of energy required to melt the cube. First, we need to raise its temperature by 10 degrees. The specific heat of ice is 2108 J /(kgK) and we have 0.01 kg of it, and so we will need
$$E_1 = 0.01 \cdot 2108 \cdot 10 = 210.8 J$$

Now, the ice needs to transition to the liquid state.
The amount of heat required to melt one kilogram of ice is 333 700 J. Thus, we need
$$E_2 = 0.01 \cdot 333 700 = 3337 J$$

So, we need the total of
$$\Delta E = 3547.8 J$$

And this energy will be sucked away from the water. So now we can sovle for the change in temperature

$$3547.8 = 0.5 \cdot 4180 \cdot \Delta T$$
$$\Delta T = 1.7$$ Degrees Celsius.

Is it the correct answer? It seems to me that the water will be way too hot.

russ_watters
Mentor
Looks right to me. My only quibble would be that you didn't include the temperature rise of the melted ice, but at this mass ratio it is just small enough not to affect your answer.*

As a gut check, from memory it takes about 80x as much energy to melt ice as to change water's temperature by 1 degree. You have 50x as much water as ice, so that's 1.6C.

* Oops, on re-checking, it is enough to matter just a little bit.

Last edited:
ChessEnthusiast
So far, so good. You have the melted ice (A.K.A. water) at 0C in the same container as the original water at 98.3C.

ChessEnthusiast and Delta2

Staff Emeritus