Thermodynamics problem - need to find amount of water that boils

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Homework Help Overview

The discussion revolves around a thermodynamics problem involving the transfer of heat between molten lead and water. The original poster seeks to determine the amount of water that boils when lead is poured into it, given specific masses and temperatures.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the application of heat transfer equations, including Q = mcΔT and the concept of latent heat for phase changes. There are inquiries about the specific heat of lead and water, as well as the heat of fusion and vaporization constants. Some participants express confusion about the calculations and the assumptions made regarding the final states of the lead and water.

Discussion Status

The discussion is ongoing with various approaches being explored. Participants are attempting to clarify their calculations and assumptions, particularly regarding the energy balance between the lead and water. There is no explicit consensus, but some guidance has been offered regarding the need to account for all heat movements and the final equilibrium state.

Contextual Notes

Participants note the importance of specific constants and the initial conditions provided in the problem. There is a focus on ensuring that the calculations reflect the correct phase changes and temperature transitions for both substances involved.

sweetpete28
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Thermodynamics problem -- need to find amount of water that boils

Suppose molten (liquid) lead, mass = 10.28 kg, is at its melting point. The lead is poured into water of mass = 658 g and initial temperature T = 18.7 degrees C. Find the amount of the water that boils. Assume no heat loss.

No clue on how to begin other than I know that Q = mcΔT...please help.
 
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Well the energy from the molten liquid will be the sum of the heat needed to cause the temperature of the water to rise to boiling point and the energy needed to change the phase of the water.

So you know that Q=mcΔT will be used for the energy needed to raise the temperature.

But what is the equation for the heat required to change the phase of the water? (what kind of 'heat' is it called?)
 


Heat required to change the phase of the water is the Latent heat...right? Equation is Q = mL where L = 2260 kJ/kg. Right?
 


Ok...um...how is this Wrong??

Heat loss by lead = (10.28)(130)(327.5-100) = 304031 J

Heat gain by water = (.658)(4186)(100-18.7) = 223931.7444 J

Mass of water turned to steam = Q / L = (304031 - 223931.7444) = 80099.25556 J = 80.09925556 kJ / 2260 kJ/Kg = .035 kg...please explain how this is wrong...thanks.
 


Can someone please advise what I am doing wrong?
 


Do I need to sue 2010 as specific heat of water?
 


Please advise on what I am doing wrong.
 


What's the heat of fusion of lead? What's the heat of vaporization of water? What's the specific heat of liquid water? You need to have all these constants on hand.

In what state does the lead begin? What energy is released when it turns from liquid to solid at its melting point?
 


ok...

Energy released when lead turns from liquid to solid at melting point:
(10.28 kg) x (25kJ/Kg) = 257kJ

Heat to raise water to 100 degrees Celsius:
(.658 kg) x (4186) x(100-18.7) = 223.9317kJ

Heat to change water to steam:
(.658) x (2260) = 1487.08kJAm I missing anything now?
 
  • #10


What temperature will the lead be at when it turns from liquid to solid? Suppose that only some of the water is boiled away. What will be the final temperature of everything that remains at the end?
 
  • #11


100 degrees Celsius?
 
  • #12


gneill said:
What temperature will the lead be at when it turns from liquid to solid? Suppose that only some of the water is boiled away. What will be the final temperature of everything that remains at the end?

sweetpete28 said:
100 degrees Celsius?

Which question did you answer?
 
  • #13


Both...but I guess that is not right. Final temp for everything that remains is 100?
 
  • #14


What is the equation i need to set up?

There is energy released when lead turns from liquid to solid, there is the heat needed to raise water to 100 degrees celsius, there is heat needed to change water to steam, there is heat lost by l;ead
 
  • #15


Do I need to use the specific heat of lead?
 
  • #16


Is the specific heat of lead needed to calculate the heat loss of the lead in addition to the energy released from the phase change (change from liquid to solid)?
 
  • #17


Evidently not. One last chance to get this right...little help please on needed equation.
 
  • #18


Please, don't multipost.

As for the question,

The heat released by lead = The heat required by water to reach 100 degrees + the heat required for it to vaporize.

Now, you don't know what the mass of water is, that's what you need to find out.


sweetpete28 said:
ok...

Energy released when lead turns from liquid to solid at melting point:
(10.28 kg) x (25kJ/Kg) = 257kJ

Heat to raise water to 100 degrees Celsius:
(.658 kg) x (4186) x(100-18.7) = 223.9317kJ

Heat to change water to steam:
(.658) x (2260) = 1487.08kJ


Am I missing anything now?

How did you get that mass to be 0.658 in that equation there then? That's the mass of lead...
 
  • #19


.658 Kg is the mass of the water; it's given in the problem.

How about this:

Energy released when lead turns from liquid to solid = (10.28)(25kJ/Kg) = 257,000J

257000 + (10.28)(130)(327 - T) = (.658)(4186)(T - 18.7)

T = 182.2411368 degrees Celsius

Heat released by lead = 257000J + 193455.7448J = 450455.7448 J

Heat required by water to reach 100 degrees C = (.658)(4186)(100-18.7) = 223931.7444J

Q = Heat required by water to vaporize

450455.7448 = 223931.7444 + Q

Q = 226524.0004J = 226.524 kJ

m = 226.524 kJ / 2260kJ/Kg = .100 kg

So the answer is .100 Kg...yes? No?
 
  • #20


sweetpete28 said:
.658 Kg is the mass of the water; it's given in the problem.

How about this:

Energy released when lead turns from liquid to solid = (10.28)(25kJ/Kg) = 257,000J

257000 + (10.28)(130)(327 - T) = (.658)(4186)(T - 18.7)

T = 182.2411368 degrees Celsius

Whoops, I inverted the masses.

And no, this isn't correct. The T you get is 182.25. Now if this were the final temperature of water, it would be waaay above boiling point, and no latent heat was considered.
 
  • #21


Ok...what about this:

257000 + 303362.8 = 223931.7444 + Q

Q = 336.4310556 kJ

m = 336.4310556 / 2260 kJ/kg = .148 kg
 
  • #22


Yep, the equation framing sounds correct. :smile:
 
  • #23


sweetpete28 said:
Is the specific heat of lead needed to calculate the heat loss of the lead in addition to the energy released from the phase change (change from liquid to solid)?

Yes. You have to account for all the heat movements and their temperature changes (if any), and it's best to go about it in an organized fashion.

Since the question asks for how much of the water boils away, it would be a reasonable guess that the final state of the system will have some liquid water left at 100C (an assumption that can be checked as the energy calculations proceed; if no contradictions arise then the assumption is confirmed). 100C then would have to be the final temperature of the system when it reaches equilibrium.

For each of the substances involved make an ordered list all the states / heat changes that must take place for it to reach the end state. So for example, water:

Start: Liquid @ 18.7C
1. Heat to 100C (boiling point)
2. Boil some of the water away @ 100C
End: Some liquid @ 100C

Make a similar list for the lead. Determine where heat is coming from (the source driving the system) and where it's going to (what is 'sinking' the heat). If you know the initial and final state (phase and temperature) of the heat source as you do in this case, you should be able to tabulate all the heat that it will supply in going from starting state to finish state. Then it becomes a matter of applying that heat to the other substance to drive its transitions from start to finish.
 
  • #24


FYI: Below answer is incorrect. Please help! What am I doing wrong now?

257000 + 303362.8 = 223931.7444 + Q

Q = 336.4310556 kJ

m = 336.4310556 / 2260 kJ/kg = .148 kg
 
  • #25


Are we sure there is a phase change for the lead? If so, is it definitely the full amount?
 
  • #26


wait wait wait...

If we start with .658 kg of water and .148 vaporizes...what is the amount of water that boils?
 
  • #27


sweetpete28 said:
FYI: Below answer is incorrect. Please help! What am I doing wrong now?

257000 + 303362.8 = 223931.7444 + Q

Q = 336.4310556 kJ

m = 336.4310556 / 2260 kJ/kg = .148 kg

Do they want the result specified in any particular units?
 
  • #28


Yes -- kilograms, so that is not the issue. Specific question is: Find the amount of water that boils. But below question it also states "Note: reaching the boiling point is not enough, the question asks for the amount of water that vaporizes as well" and to "assume that no water evaporates, it only vaporizes at the boiling point"
 
  • #29


sweetpete28 said:
Are we sure there is a phase change for the lead? If so, is it definitely the full amount?

You should be able to answer that question by determining how much energy it would take to boil away all the water starting from the water's initial temperature. Compare to the energy released when all the lead changes from liquid to solid. If more energy is required than is available from the lead phase change, then all of the lead will change.
 
  • #30


sweetpete28 said:
Yes -- kilograms, so that is not the issue. Specific question is: Find the amount of water that boils. But below question it also states "Note: reaching the boiling point is not enough, the question asks for the amount of water that vaporizes as well" and to "assume that no water evaporates, it only vaporizes at the boiling point"

Well, your result is at least in the right ballpark then. Are the constants that you're using for the various heats ones given by the book or are they found elsewhere? (could be an accuracy issue).
 

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