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

[Shmiam]

Homework Statement

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.

Homework Equations

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

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

 

[haruspex]

I'm guessing it'd be the volume of steam which will share the same final T

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?

 

[Shmiam]

What must happen before any water boils off?

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.

 

[haruspex]

The water needs to rise to 100C first, right? I'm just generally confused how to work the equation up to that point,

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

 

[Shmiam]

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

Do they both end up at 100C? So the horseshoe would drop 1100C? But then for the heat absorbed by the water, I am 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 that's how much heat the water absorbs?

 

[haruspex]

Do they both end up at 100C?

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?

 

[Shmiam]

how much heat has the water taken up at this point?

Would that be = mc(70C)?

 

[haruspex]

Would that be = mc(70C)?

Maybe - you have not defined m and c.

 

[Shmiam]

Maybe - you have not defined m and c.

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

 

[haruspex]

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

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

 

[Shmiam]

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

Would its temperature change also equal 70C?

 

[haruspex]

Would its temperature change also equal 70C?

I didn't ask that.
You calculated that the water absorbed mc (70C) of heat. So how much did the horseshoe lose in that time?

 

[Shmiam]

So how much did the horseshoe lose in that time?

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

 

[haruspex]

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

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?

 

[Shmiam]

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?

I'd have to say I don't know

 

[haruspex]

I'd have to say I don't know

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

 

[Shmiam]

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

Oh, then the horseshoe must have lost Q

 

[haruspex]

Oh, then the horseshoe must have lost Q

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

 

[Shmiam]

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

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

 

[haruspex]

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

Ok. Time to turn that into a numerical temperature.

 

[Shmiam]

Ok. Time to turn that into a numerical temperature.

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..

 

[haruspex]

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..

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)).

 

[TJGilb]

Homework Statement

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.

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).

 

[Shmiam]

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)).

In that case, the horseshoe loses 454.567K

 

[haruspex]

In that case, the horseshoe loses 454.567K

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

 

[Shmiam]

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

The water goes through its phase change from liquid to steam, and it stops if it all evaporates or if the system reaches thermal equilibrium

 

[haruspex]

The water goes through its phase change from liquid to steam, and it stops if it all evaporates or if the system reaches thermal equilibrium

Good.
How much heat does the horseshoe still have to lose for equilibrium?

 

[Shmiam]

Good.
How much heat does the horseshoe still have to lose for equilibrium?

At equilibrium it would match the temperature of the water, so Q = MC(373K-1018.433K), which comes out to be -208076.96J?

 

[haruspex]

At equilibrium it would match the temperature of the water, so Q = MC(373K-1018.433K), which comes out to be -208076.96J?

Right, so how much water will boil off.

 

[Mister T]

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?

If there were no water boiling off, how would you use $Q=mc\Delta T$ to find the final temperature?

 

[Shmiam]

Right, so how much water will boil off.

with delta(E) = 0 = Qshoe + Qsteam
-Qshoe = Qsteam
208076.96J = L Msteam with L being the latent heat of vaporization = 2256 kJ/kg = 2256000 J/kg
Msteam = .0922kg = .0922L

 

[haruspex]

with delta(E) = 0 = Qshoe + Qsteam
-Qshoe = Qsteam
208076.96J = L Msteam with L being the latent heat of vaporization = 2256 kJ/kg = 2256000 J/kg
Msteam = .0922kg = .0922L

Looks right.

 

[Shmiam]

Looks right.

Awesome! Thank you very much for all of the guidance, I think it helped me to understand the concepts here much more

 

[haruspex]

Awesome! Thank you very much for all of the guidance, I think it helped me to understand the concepts here much more

That's a good result then.