# Hot waterr mixed with ice

#### chawki

1. The problem statement, all variables and given/known data
Hot water is mixed with an amount of ice having an equal mass to that of the water
and an initial temperature of 0 °C. The specific heat capacity of
water is 4.2 kJ/(kg K) and the specific latent heat of fusion for ice is 334 kJ/kg. Assume that no
heat is lost to the surroundings.

2. Relevant equations
What should the initial temperature of the hot water be to achieve a final water temperature of 5 °C with all the ice melted?

3. The attempt at a solution
Q=m*Hf
Q=m*Cp*delta T

m*334 = m*4.2*(5-T)
(5-T) = 334/4.2
T = -74 C (i'm very probably wrong :grumpy:)

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#### gneill

Mentor
The ice at 0C melts to water, then rises in temperature to 5C. Meanwhile, the energy to accomplish this came from hot water at some initial temperature (call it T) that cooled to 5C.

#### chawki

I'm still wondering how to write that in equation

#### gneill

Mentor
I'm still wondering how to write that in equation
Break it down into the steps I outlined to write the equation. Start with the ice melting. How much heat is required? Then raise the temperature of the melted water to 5C. How much heat is required?...

#### chawki

The heat required for ice melting is Q=m*Hf = 334m J ?

#### gneill

Mentor
Assuming appropriate units for the mass m, yes. Remember, the heat of fusion is given in units of kilojoules per kilogram.

Now that the ice has been melted, how much energy will it take to raise its temperature to 5C?

#### chawki

to raise ?
i got the thought that it's gonna be like this:
334m=m*4.2*(T-5)
334=4.2*t-21
4.2t=355
t=84.52 C ?

#### gneill

Mentor
Can you explain what each of the terms in your equation represents physically? For example, the 334m represents the heat required to melt a mass m of ice. What precisely does the RHS of that equation represent?

i'm wrong ?

#### gneill

Mentor
Not so much wrong as incomplete. When you write out a heat balance equation, you have to keep in mind all the things that need heating or cooling, detailing where all the energy comes from and goes to.

So far you've showed that you understand the heat required to melt the ice. Melting ice yields liquid water. What temperature is that water?

i'm lost...
334*m = 4.2*m*T
T = 79.52

#### gneill

Mentor
If you enumerate all the things that have to happen to reach the goal, you will find that things become clear.

1. Ice has to melt. That takes 334 kJ/kg. It turns into water at 0C. No ΔT for this.
2. The water at 0C has to be warmed to 5C, the final temperature. ΔT is +5C. Cp = 4.2 kJ/kg/K.

Where does the energy to do this come from? The initially hot water at some as yet unknown temperature T. So,

3. The hot water at T is cooled down to the final temperature. ΔT = T - 5C. Cp = 4.2 kJ/kg/K.

So the heat represented in steps 1 and 2 must equal that in step 3.

#### chawki

but since you say ''The hot water at T is cooled down to the final temperature'' which is 5 C.
it means that ΔT=5-T and not the reverse

#### gneill

Mentor
but since you say ''The hot water at T is cooled down to the final temperature'' which is 5 C.
it means that ΔT=5-T and not the reverse
If you wish. I was just enumerating the magnitudes of the heat energies involved. If you put everything on the appropriate sides of the "equals", you can deal entirely with positive values.

#### chawki

ok so it's as i wrote in post #1 ? but then the temperature is minus :(

#### gneill

Mentor
Identify each of the individual heats associated with the steps I enumerated in post #12 above in the equations of your post #1.

#### chawki

ok well, there is TWO ΔT, the one of ice going from 0C to 5C
and the one of hot water going from T to 5C. right ?

#### chawki

the Q=334*m is the Q of ice, right?
then the Q of hot water is Q=m*4.2*(5-T) right ?

#### gneill

Mentor
the Q=334*m is the Q of ice, right?
then the Q of hot water is Q=m*4.2*(5-T) right ?
Right. And there's one more ΔT associated with the formerly-ice going from 0C to 5C. As I said, enumerate all the things that have to happen to get to the goal and identify the heats associated with them.

#### chawki

Right. And there's one more ΔT associated with the formerly-ice going from 0C to 5C. As I said, enumerate all the things that have to happen to get to the goal and identify the heats associated with them.
ok so for ice there is two types of energy ?
Q=m*334
and
Q=m*4.2*(5-0)

#### gneill

Mentor
ok so for ice there is two types of energy ?
Q=m*334
and
Q=m*4.2*(5-0)
It's all heat energy. And once it's melted, it's no longer ice but water at 0C. The heat of fusion is the heat energy required to turn ice at 0C to water at 0C. After that you have to treat it as plain liquid water with the usual Cp.

#### chawki

i appreciate your help really.... but ahhhhhhhhhhhh why it has to be complicated
so we have 3 equations, right?

#### gneill

Mentor
Well, one equation relating all the steps, balancing all the heat energy as it moves from place to place. I've identified for you the three steps that have to happen. Each has an amount of heat involved. You should always be able to identify the steps required in any of these sorts of problems and write an expression for the heat involved.

Look, perhaps an analogy will help you. Imagine that you've got a hole in the ground that's 9.6cm deep (never mind the circumference of the hole for this discussion), and you want to fill it and leave a mound of dirt 5cm high on top of it. Beside the hole is a mound of dirt you have to use.

First you have to fill the hole (melt the ice). Then you have to pile 5cm of dirt on top of the filled hole (raise the temperature to 5C). If the height of the depleted mound is now even with the new 5cm mound on top of the hole, how high was the original mound? That's a 'physical equivalent' for your heat problem. Perhaps it will help you to picture the steps.

#### chawki

ok well. the equations are :
Q1=m*334
Q2=m*4.2*(5-T)
Q3=m*4.2*5

#### chawki

ok well. the equations are :
Q1=m*334
Q2=m*4.2*(5-T)
Q3=m*4.2*5
should we add Q2 and Q3 together?

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