# Homework Help: What is the final temperature?

1. Jul 18, 2013

### shin777

30g of ice at -15 degree celcius is added to 100g of water at 25 degree celcius. What is the final temperature?

T1=ice
T2=H2O

at equilibrium must have
m1c1(Te-T1)+m1k=m2c2(T2-Te) [k=3.3x10^5 J/K fusion heat coeff.]

then
Te=(m1c1T1+m2c2T2-m1k)/(m1c1+m2c2)
T1=268 k
T2=298 K
c1=2090 J/(kg K)
c2=4186 J/(kg K)
Te=(0.02*2090*268+0.1*4186*298-0.02*3.3… / (0.02*2090+0.1*4186)
Te=280,9 K
Te=7.8 C

Final Temperature is 7.8 C

does this look ok?

2. Jul 18, 2013

### PeterO

Specific heat for ice is about half that of water, so you will need much less heat to warm the ice up to zero degrees than it takes to cool the water to zero degrees.
Less mass of ice, smaller temperature change and easier to change so a win on all three counts.

Big question - how much heat do you need to melt the ice?

If you need more heat to melt the ice than you need to cool the water to zero degrees you will end up with an ice/water mix at zero degrees.

3. Jul 18, 2013

### shin777

hmm.. my tutor help me to solve this problem but now I look at it again, some of these equation don't make sense. any suggestion? :(

4. Jul 18, 2013

### SteamKing

Staff Emeritus
How many kg is 30 g of water?

5. Jul 18, 2013

### SteamKing

Staff Emeritus
Rather than write one huge equation which no one can understand, and which is prone to error, look at the problem in stages:

1. Find the amount of heat transfer to the ice to bring its temp. to 0 deg. C
2. Find the amount of heat transfer to the ice to cause its phase change from solid to liquid.
3. Find the equilibrium temperature of the now melted ice and the cooled water.

In this way, it is easier to detect and fix any mistakes in the calculations.

6. Jul 18, 2013

### CWatters

There doesn't appear to be enough energy in the starting components for all 130g to be water.

One way to do this is to work out the energy contained in the starting components (eg relative to absolute zero). Then apply conservation of energy... pretend you have 130g of ice at abs zero and see how hot it would end up be if you added that energy.

There appears to be enough to raise it all to 273K (aka 0C) but not quite enough left to melt it all to water.

Last edited: Jul 18, 2013
7. Jul 18, 2013

### shin777

latent heat of fusion(L) = 80 cal g
Mc = mass of ice

heat gained = (Mc *4180 * 15) + (Mc * L) + (Mc * 4180 * t)
heat gained = (30 * 4180 *15) + (30 * 80) + (30 * 4180 * t)
t = -15c

final temp is -15c

Last edited: Jul 18, 2013
8. Jul 18, 2013

### CWatters

9. Jul 18, 2013

### PeterO

If you are going to use 80 for the latent heat, you need to use 1 for the SHC of water and about 0.5 for the ice.*

To heat the ice to 0 degrees you need

0.03 x 15 x 2090 Joules.

To melt all that ice you need 0.03 x 3.3 x 105 J

So to get the ice from -15 up to water at 0 degrees you need the sum of those two.

To Cool the water to 0 degrees you need to extract

0.1 x 25 x 4180 Joules

That is more than enough to get the ice up to zero degrees, but nowhere near enough to melt all the ice - so you will be left with a mix of ice and water - naturally at zero degrees.