Homework Help: How much water boils off when a hot horseshoe is dropped in

1. Dec 2, 2016

Shmiam

1. The problem statement, all variables and given/known data
A blacksmith drops a 0.7 kg horseshoe of iron at a temperature of 1200 ̊C into a bucket of
0.5 L
of water at an initial temperature of 30 ̊C. How much of the water boils off? Assume the bucket
absorbs none of the heat.

2. Relevant equations
I believe Q = mc delta(T) is what I'm aiming to use here, but I'm not entirely sure how

3. The attempt at a solution
I'm a bit lost, I know that the two objects (shoe and water) will ultimately reach the same final temperature, and I can write Q = mc delta(T) to reflect that, but I'm unsure how it corresponds to the water that boils off, and how I might find that value? I'm guessing it'd be the volume of steam which will share the same final T as the water and shoe? Not sure how to find it though.

Thank You

2. Dec 2, 2016

haruspex

No, the steam will come of at 100C and take no further part.
Think of it in stages, and assume that at any given moment all the water is at the same temperature.
What must happen before any water boils off?

3. Dec 2, 2016

Shmiam

The water needs to rise to 100C first, right? I'm just generally confused how to work the equation up to that point, and then what I need to do to find how much water boils off.

4. Dec 2, 2016

haruspex

Yes.
What heat will the water have absorbed? What temperature drop occurs in the horseshoe?

5. Dec 2, 2016

Shmiam

Do they both end up at 100C? So the horseshoe would drop 1100C? But then for the heat absorbed by the water, im not sure what it equals. Is it related to the horseshoe's Q = mc delta(T)? As in, I solve that with delta(T) = 1100C and thats how much heat the water absorbs?

6. Dec 2, 2016

haruspex

Eventually, perhaps, but at this stage we are just considering the process up to the point where all the water is at 100C. The horseshoe may still be hotter.
Answer my first question: how much heat has the water taken up at this point?

7. Dec 2, 2016

Shmiam

Would that be = mc(70C)?

8. Dec 2, 2016

haruspex

Maybe - you have not defined m and c.

9. Dec 2, 2016

Shmiam

c is the specific heat of water and m is the mass, which, using density, would be .5kg

10. Dec 2, 2016

haruspex

OK, so how much heat has gone from the horseshoe?

11. Dec 2, 2016

Shmiam

Would its temperature change also equal 70C?

12. Dec 2, 2016

haruspex

You calculated that the water absorbed mc (70C) of heat. So how much did the horseshoe lose in that time?

13. Dec 2, 2016

Shmiam

What I meant by that was whether the horseshoe lost mc (70C) or if thats the wrong way to think about it

14. Dec 2, 2016

haruspex

You need to base your thinking on sound principles, not wild guesses.
It is a very simple question: if the water gained a quantity Q of heat from the horseshoe, what quantity of heat did the horseshoe lose to the water?

15. Dec 2, 2016

Shmiam

I'd have to say I don't know

16. Dec 2, 2016

haruspex

Heat is energy. Energy is conserved. If the total stays the same, but the water gained Q, how much did the horseshoe lose?

17. Dec 2, 2016

Shmiam

Oh, then the horseshoe must have lost Q

18. Dec 2, 2016

haruspex

Right. So what is the temperature of the horseshoe at this point?

19. Dec 2, 2016

Shmiam

So, setting the two Q's equal, I found delta(T) of the shoe to be = -(mc 70C)/(MC)
m - mass of the water
c - specific heat of water
M - mass of the shoe
C - specific heat of iron

20. Dec 2, 2016

haruspex

Ok. Time to turn that into a numerical temperature.

21. Dec 2, 2016

Shmiam

I think I must be missing something here
m = .5kg
c = 4187 J/(kg*K)
M = .7kg
C = 460.548 J/(kg*K)
and 70C + 273K = 343K

I'm getting an outrageously high change..

22. Dec 2, 2016

haruspex

You don't need to convert to K. A change of x degrees in C is also a change of x degrees in K, and here we are only concerned with temperature changes.
The heat gained by the water is mc(100-30) or, equivalently, mc((273+100)-(273+30)).

23. Dec 3, 2016

TJGilb

According to the problem you know that no energy is lost from the system, therefore you can say that the change in energy of zero is equal to the sum of every Q. The Q you'll have to account for include the heat loss from the horseshoe, the heat gained to bring the water up to 100C, and the heat gained to evaporate the water. Heat exchange will continue until either the horseshoe is removed, or they come to thermal equilibrium. If you assume that not all of the water will evaporate, this tells you the final temperature of the horseshoe, which should then allow you to solve for its Q (if this turned out to be an incorrect assumption, you would be able to tell because you would find that all of the water evaporated, but then that would also be your answer).

24. Dec 3, 2016

Shmiam

In that case, the horseshoe loses 454.567K

25. Dec 3, 2016

haruspex

Right. And now we have all the water at 100C.
What process will happen next, and what will eventually halt that process?