How long does it take to boil water in a 600 watt microwave oven?

Boiling is very definitely a spontaneous process. For example, submerge a vessel containing water in a vacuum chamber. As you reduce the pressure, the water will boil because the vapor pressure of the water at room temperature is effectively greater than the ambient pressure. No energy input is required.In summary, the conversation discusses estimating the time it takes to bring a cup of water to boiling temperature in a microwave oven. The equations P=W/t and Q=W=m*c*delta(T) are introduced, and an attempt is made to solve for the time using these equations. The concept of work versus heat is discussed, with the conclusion that heating water involves heat transfer, not work. It is also noted that boiling water is a spontaneous
  • #1
pentazoid
146
0

Homework Statement




Estimate how long it should take to bring a cup of water to boiling temperature in a typical 600 watt microwave oven, assuming all energy ends up in the water(assume any reasonable initial temperature for the water.) What can you conclude about the amount of heat added to the water?

Homework Equations



P=W/t
possibly Q=W=m*c*delta(T)

The Attempt at a Solution



P=600 watts
W=Q=m*c*delta(T)
c=4.2 J/kg*1
m=.250 grams ( I guessed what the mass of the cup of water would be)
W=(.250)(4.2)(73 K)=76.7 Joules
t=W/P=76.7 joules/600 watts

W/P=t
T_initial=300 K

T_final=373 K
c=4.2
delta(T)=T_final-T_initial=373 K - 300 K= 73 K
not sure if I should just guesstimate the mass of the cup of water.

W=(.250)(4.2)(73 K)=76.7 Joules
t=W/P=76.7 joules/600 watts

Boiling the cup of water in the cup involves work not heat. Energy must be leaving the system(water in this case) to be described as heat. If the cup of water was cooling off, then I could describe that process as heat, not work.
 
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  • #2
It looks like you have the right idea. However, the mass of water will be a lot more than 0.250 g.

Can you come up with a better estimate for m? Perhaps first you should estimate the volume first.
 
  • #3
Redbelly98 said:
It looks like you have the right idea. However, the mass of water will be a lot more than 0.250 g.

Can you come up with a better estimate for m? Perhaps first you should estimate the volume first.

I actually meant to write the mass of the cup was supposed to be .250 kg , not .250 grams. Is my explanation for why there is no heat involved in the process correct?
 
  • #4
Well, no. There is heat involved. The microwaves are a means of adding heat to the water. In fact, the sole function of a microwave oven is to heat water.

Doing work would involve a change in volume, or some kind of displacement, since
W = P ΔV .or. F d cosθ​

I guess on a microscopic level, heating something is equivalent to doing work on many individual molecules all at once. But we don't really consider that as work in a practical sense.
 
  • #5
pentazoid said:
Energy must be leaving the system(water in this case) to be described as heat.

This isn't true; any transfer of thermal energy is described as heat. The water is being heated through irradiation; its thermal energy increases.
 
  • #6
Redbelly98 said:
Well, no. There is heat involved. The microwaves are a means of adding heat to the water. In fact, the sole function of a microwave oven is to heat water.

Doing work would involve a change in volume, or some kind of displacement, since
W = P ΔV .or. F d cosθ​

I guess on a microscopic level, heating something is equivalent to doing work on many individual molecules all at once. But we don't really consider that as work in a practical sense.

Heat is the spontanteous flow of energy from one object to another and the amount of energy that enter a system . When spontaneous energy is involved, energy usually flows from a higher state of energy to a lower state of energy. hot objects have higher thermal energy than cold objects do. When we rub our hands together and they get hotter, heat is not involved. When cup of water at room temperature temperature goes up , its because of work not heat.
 
  • #7
pentazoid said:
Heat is the spontanteous flow of energy from one object to another and the amount of energy that enter a system .

This also describes work.

pentazoid said:
When cup of water at room temperature temperature goes up , its because of work not heat.

You can keep asserting this, but those responding to your question disagree. More importantly, in the water/cup system you can't identify any significant examples of work, whether it's force + change in distance, pressure + change in volume (other than routine thermal expansion), electric charge + change in voltage, etc. I invite you to try.

The microwaves carry entropy, which is the defining characteristic of heat transfer.

During thermal expansion and boiling, the water and gas does do work on the atmosphere; is this what you meant?
 
  • #8
Mapes said:
This also describes work.



You can keep asserting this, but those responding to your question disagree.

According to Schroeder,my textbook author, heat is "defined as any spontaneous flow of energy caused by a difference in temperature". He then goes on to give examples of heat energy transfer , and none of the examples described energy flowing into a hotter region of space. One example says that when you rubbed your hands together , heat is not responsible for you hands getting warmer because the rubbing of your hands is not a spontaneous process because when you rubbed your hands together , you are producing a higher state of energy. Here is a couple of definitions of heat from wikipedia:

Heat may be defined as energy in transit from a high-temperature object to a lower-temperature object." ( ^ Discourse on Heat and Work - Department of Physics and Astronomy, Georgia State University: Hyperphysics (online))






During thermal expansion and boiling, the water and gas does do work on the atmosphere; is this what you meant?

Yes work cause water to boiled to a certain temperature , but boiling water is not a spontaneous process , and therefore boiling water is not a heat induced process.
 
  • #9
OK, I think I see where the problem lies. You're concluding that the process of water getting hotter in a microwave over can't be classified as heating because the physical microwave oven is cooler. But your textbook example of heat transfer from a hotter to a colder body assumes that the bodies are each in thermodynamic equilibrium (i.e., they each have a single, well-defined temperature). A microwave oven is not in thermodynamic equilibrium. And I stand behind my earlier points that entropy transfer is associated with heat, not work, and that none of the standard generalized force-displacement pairs can be connected to water getting warmer in a microwave oven.

pentazoid said:
boiling water is not a spontaneous process , and therefore boiling water is not a heat induced process.

Of course it's spontaneous above 100°C! Seriously, I think you should consider your arguments more carefully.
 
  • #10
Radiation is a well-known, and scientifically accepted, means of transferring heat.
http://en.wikipedia.org/wiki/Heat_transfer#Radiation

pentazoid said:
When cup of water at room temperature temperature goes up , its because of work not heat.

What if the water is in a black cup and gets heated by direct sunlight?

Even if we accepted your definition of heat transfer, and your arguments that heating is not taking place, you have not shown that doing work is taking place either.
 
  • #11
Redbelly98 said:
Radiation is a well-known, and scientifically accepted, means of transferring heat.
http://en.wikipedia.org/wiki/Heat_transfer#Radiation



What if the water is in a black cup and gets heated by direct sunlight?

Even if we accepted your definition of heat transfer, and your arguments that heating is not taking place, you have not shown that doing work is taking place either.

Work , according to Scroeder is the transfer of energy into or out of a system.(Scroeder, p.19 )
Energy of a system will increase and therefore its temperature will increase. The temperature of the cup of water reaching its boiling pt. Therefore its temperature changes, therefore there is work being done on the cup. I don't know , maybe Scroeder did not defined heat properly .

You do make a good point since radiation is a type of heat transfer , and microwaves are part of the EM spectrum and excited the water molecules inside the microwave.
 
  • #12
No, these are not good definitions, because heat is also the transfer of energy into or out of a system. Jeez, even Wikipedia is more precise. A good distinguishing characteristic is whether entropy is transferred (heat) or a mechanical variable (displacement, volume, charge, magnetization, strain, etc.) (work).
 
  • #13
Mapes said:
No, these are not good definitions, because heat is also the transfer of energy into or out of a system. Jeez, even Wikipedia is more precise. A good distinguishing characteristic is whether entropy is transferred (heat) or a mechanical variable (displacement, volume, charge, magnetization, strain, etc.).

Well take it up with the author of my thermal physics textbook not me. I didn't right these definitions for heat. We are at a point in the class where we have not covered entropy. But from what I can remember what definition of entropy is from another class, entropy of a system depends on the number of arrangements that system can have or its multiplicity.
 
  • #14
pentazoid said:
Work , according to Scroeder is the transfer of energy into or out of a system.(Scroeder, p.19 )

Defining work as "the transfer of energy" would mean it is not mutually exclusive of adding/removing heat, and that heat transfer is one way to do work.

But from what I remember of thermodynamics, heat transfer and doing work are separate, distinguishable processes for changing the energy content of a system:
ΔE = W + Q​
gives the change in energy of a system, where W is work done on the system, and Q is the heat added to it.
 
  • #15
At this point, it's probably best to wait and see what your professor says about all this.
 
  • #16
Back to the original question ...

pentazoid said:
I actually meant to write the mass of the cup was supposed to be .250 kg , not .250 grams.

That's okay. However, note that c=4.2 J/g/K, not 4.2 J/kg/K as you wrote before.
 
  • #17
Redbelly98 said:
Back to the original question ...



That's okay. However, note that c=4.2 J/g/K, not 4.2 J/kg/K as you wrote before.

My estimate came out to be near 10 minutes. Is that a reasonable time?
 
  • #18
10 minutes sounds a little long. Can you show your calculation?
 
  • #19
Redbelly98 said:
10 minutes sounds a little long. Can you show your calculation?

yes. t=Q/P=(250)*373*(4.2)/600 =652 seconds= almost 11 minutes.
 
  • #20
Earlier you calculated the temperature difference as 73 K, not 373 K.
 
  • #21
Mapes said:
Earlier you calculated the temperature difference as 73 K, not 373 K.

your right,thanks for pointing that out. I now get my time to be about 2 minutes
 
  • #22
Looks good.
 

1. What is heat transfer?

Heat transfer refers to the movement of thermal energy from one place to another. It can occur through three mechanisms: conduction, convection, and radiation.

2. What is a heat transfer problem?

A heat transfer problem is a mathematical model that describes the rate at which thermal energy is transferred between different objects or materials. It involves solving equations that govern the behavior of heat transfer in order to determine the temperature distribution and flow of heat within a system.

3. How is heat transfer problem solved?

Heat transfer problems are typically solved using numerical methods, such as finite difference or finite element analysis, which break down the problem into smaller, solvable parts. These methods involve discretizing the system into a finite number of elements, and then using equations and algorithms to solve for the temperature distribution and heat flow within each element.

4. What are some real-world applications of heat transfer problems?

Heat transfer problems are used in a wide range of industrial and engineering applications, such as designing heat exchangers, thermal insulation, and HVAC systems. They are also important in understanding natural phenomena, such as heat transfer in the Earth's atmosphere and oceans.

5. What are some challenges in solving heat transfer problems?

One of the main challenges in solving heat transfer problems is accurately modeling the complex physical processes involved, such as convection and radiation. Additionally, the boundary conditions and material properties used in the model must be carefully chosen and may vary significantly depending on the specific application. Another challenge is the computational complexity of solving these problems, particularly for large or highly nonlinear systems.

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