Do you need to account for the water heating up to the boiling point?

[curly_ebhc]

High School Physics Lab: Take 200mL of water (Room temp) and place it in a microwave on high for 60 seconds. Calculate the Energy transferred to the water by the microwave.
Pretty easy:
Step 1: Heat of Temp Change : Q= mC∆T where m=200mL
Step 2: Add Heat due to phase change: Q=mL where m= mass lost
QUESTION: The question is do you need to account for the water heating up to the boiling point?

The physics department has two opinions on this.
Opinion 1: The water must heat up to 100 ℃ before it can undergo phase change. Therefore Step 2 should include a mC∆T + mL to account for the temperature change for the mass of water lost.

Opinion 2: Due to the fact that temperature describes a distribution of kinetic energies, the particles that have enough energy to vaporize already have their heat energy added in Step 1.

Disclaimer: We are a bunch of high school physics teachers who once earned degrees that forced us to understand this stuff, but that was many eons ago. Also as far as the actual values in the lab go, the difference between opinion 1 and opinion 2 are negligible.

 

[phinds]

You say you are calculating the energy transferred to the water by the microwave. So, is it your belief that the microwave is transferring no energy to the water as it heats the water up to the boiling point?

 

[berkeman]

place it in a microwave on high for 60 seconds.

What is the RF output power rating of the microwave oven? (you can usually find it on the label...)

 

[gmax137]

My microwave oven will not boil 200 ml (just under 1 cup) of water in 60 seconds. So, I would say this experiment would better measure the temperature at zero and 60 seconds, then use m cp dT to calculate the energy transferred to the liquid. There wouldn't be any mass change / phase change in my oven.

 

[gmax137]

for extra credit, you could estimate the energy transferred from the heating liquid into the cup, and from the cup into the oven atmosphere. That would be energy transferred into the liquid and then transferred out before you took the 60 second temperature.

 

[curly_ebhc]

You say you are calculating the energy transferred to the water by the microwave. So, is it your belief that the microwave is transferring no energy to the water as it heats the water up to the boiling point?

There is energy being transferred to the water, both to change the temperature and change the phase of the evaporated water.
The question is really about weather the water that evaporates really reaches 100 ℃. For example: The water heats to 80 ℃ and 10 mL of water are lost. Does the 10 mL of lost water make it all the way to 100 ℃ or just 80℃ but it is the water at the upper end of the energy distribution?

What is the RF output power rating of the microwave oven? (you can usually find it on the label...)

This is really for the students to practice their equations. I sometimes do some basic error analysis with the microwave power, but I am not really concerned about the microwave's actual output power for this lab.
BTW does anyone know if the power listed on the sticker is the electrical power input into the magnetron or is it actually a measured power of microwave energy?

My microwave oven will not boil 200 ml (just under 1 cup) of water in 60 seconds. So, I would say this experiment would better measure the temperature at zero and 60 seconds, then use m cp dT to calculate the energy transferred to the liquid. There wouldn't be any mass change / phase change in my oven.

There is most definitely a phase change as some of the water evaporates. Students easily find the lost mass.

 

[berkeman]

BTW does anyone know if the power listed on the sticker is the electrical power input into the magnetron or is it actually a measured power of microwave energy?

Mine lists both...

 

[berkeman]

There is most definitely a phase change as some of the water evaporates. Students easily find the lost mass.

Does the oven that they use have a turntable? It seems strange that you would get much evaporation in that short interval with even heating. If you don't have a turntable, I suppose there could be some standing wave hotspots in the liquid that make it to 100C.

 

[gmax137]

There is most definitely a phase change as some of the water evaporates. Students easily find the lost mass.

OK, this surprises me. But I've been wrong before. Now I have to go do this myself...

 

[kuruman]

Now I have to go do this myself...

Me too.

 

[phinds]

The question is really about weather the water that evaporates really reaches 100 ℃.

Ah. Well, see, I was clearly confused because of your categorical (I thought) statement of the question:

Calculate the Energy transferred to the water by the microwave

 

[gmax137]

For example: The water heats to 80 ℃ and 10 mL of water are lost.

Are these values typical? I just tried this, twice.

First trial, 60 seconds brought the water from 68 to 153 F (20 to 67C). I saw no change in the mass.
Second trial, I ran the oven for 90 seconds to bring the water from 68 to 186F (20 to 85C). My mass went from 7.0 to 6.9 ounce.

Unfortunately all I have is a cheap digital kitchen scale, so the "loss" of 0.1 oz (~3 gram) has a lot of uncertainty. But your 10 ml should be seen even on my scale, that's 0.35 ounces.

I will have to think of a better way to do this.

OBTW, using the numbers I measured, it works out to 649 watts in trial 1 and 651 watts in trial 2 (ignoring the mass defect). At least that is consistent.

 

[jbriggs444]

If one is doing careful calorimetry -- ROFL, not in this situation, clearly. Then one should be worried about more than just the possibility of water evaporating below 100 degrees C.

What if the water ends up super-heated? Could we lose 10 ml by surface evaporation? Could we have ordinarly boiling followed by super-heating?

Could we lose 10 ml by evaporation before, during and after we pour the water out? If we are being this boneheaded with the calorimetry, goodness knows how casual we are being with everything else.

What if part of the microwave energy goes into heating the already vaporized water? Does that count?

What is the [average] temperature of the resulting vapor anyway?

Given that the water vapor will be condensing out on the microwave walls, exactly how were we planning to measure that temperature?

What about the energy lost to the container?

Apparently microwave oven size is going to be 15 liters or more. Ten ml is about ten grams is about half a mole of stuff with a molecular weight of 18. About 11 liters. So we will not be filling the microwave with steam at ambient pressure. Nor will we be heating the walls to 100 C.

 

[curly_ebhc]

OK, this surprises me. But I've been wrong before. Now I have to go do this myself...

Ok. I might have been simplifying values so I actually looked at a student's lab.
Most students only lose 2-3 grams of water for an initial 200 grams. This only accounts for about 10% of the energy absorbed by the water.

 

[curly_ebhc]

If one is doing careful calorimetry -- ROFL, not in this situation, clearly. Then one should be worried about more than just the possibility of water evaporating below 100 degrees C.

What if the water ends up super-heated? Could we lose 10 ml by surface evaporation? Could we have ordinarly boiling followed by super-heating?

Could we lose 10 ml by evaporation before, during and after we pour the water out? If we are being this boneheaded with the calorimetry, goodness knows how casual we are being with everything else.

What if part of the microwave energy goes into heating the already vaporized water? Does that count?

What is the [average] temperature of the resulting vapor anyway?

Given that the water vapor will be condensing out on the microwave walls, exactly how were we planning to measure that temperature?

What about the energy lost to the container?

Apparently microwave oven size is going to be 15 liters or more. Ten ml is about ten grams is about half a mole of stuff with a molecular weight of 18. About 11 liters. So we will not be filling the microwave with steam at ambient pressure. Nor will we be heating the walls to 100 C.

I absolutely agree with everything here.

Let us say that it was a perfect experiment. I assume you should get lost mass without boiling? Do the molecules that vaporize reach 100 C or are they just the molecules with enough energy to break the intermolecular bonds keeping them liquid?

 

[jbriggs444]

I absolutely agree with everything here.

Let us say that it was a perfect experiment. I assume you should get lost mass without boiling? Do the molecules that vaporize reach 100 C or are they just the molecules with enough energy to break the intermolecular bonds keeping them liquid?

Water can evaporate at any temperature. The resulting vapor will be at that same temperature. Below zero, we call it sublimation.

The thing that is special about "boiling" is that you can get evaporation within the bulk of a liquid (at a nucleation point) so that the resulting bubble grows rather than shrinking away to nothing.

 

[curly_ebhc]

I just wanted to say thanks to everyone who has looked at this. I have been impressed with the discussion, especially that some of you were willing to try the experiment.

 

[haruspex]

Opinion 2: Due to the fact that temperature describes a distribution of kinetic energies, the particles that have enough energy to vaporize already have their heat energy added in Step 1.

I fail to see how this produces a different answer from opinion 1. Each molecule that evaporates has to reach a certain energy to escape and carries that energy away with it. If you start with mass M and mass m evaporates the energy transferred is MC∆T+mL.

If I pour a glass of water from the tap and leave it to stand (humidity being less than 100%) it will gradually evaporate. In the process it will get colder and absorb heat from the surrounding. Even if I perfectly insulate it, it will do that until its temperature drops to the dew point. Once there, evaporation matches condensation.

I suspect that the state of boiling is widely misunderstood. It is not that all the molecules simultaneously reach the energy required to escape. Rather, the temperature reaches the point at which the SVP matches the atmospheric pressure, allowing bubbles of vapour to form within the water. This greatly increases the surface area, allowing the rate of vaporisation to match the rate of heat input.

 

[hmmm27]

200ml ~40C water in a 250ml Pyrex kitchen measuring cup,
consumer turntable microwave.
no thermometer.

120s : 2 minutes

result : interior air noticeably warmer and humid
water now ouch hot ~60C ?,
lost maybe 3-4ml.

Not very dramatic : are students doing this at home ?

 

[kuruman]

Why not switch to vegetable oil? It could be a bit messy but it doesn't have phase transition problems and it's cheap. Just a thought $\dots$

 

[berkeman]

water now ouch hot ~60C ?,

Reminds me of the EE test for whether a power component is running too hot...

Spoiler

If you can't keep your finger on the component or the heat sink, the component is running too hot with the junction temperature exceeded.

 

[erobz]

Some other thoughts:

1) On most microwaves you can adjust the output power (time averaged power) programmatically. look for a PWR button or PWR LEVEL, etc..., input number 1-10 (or percentage).

2) Start with water from an ice bath at $0 ^{\circ}C$, instead of room temp.

3) Consider shortening the time duration(maybe $30~\rm{s}$)

 

[Lnewqban]

Reminds me of the EE test for whether a power component is running too hot...

Spoiler

If you can't keep your finger on the component or the heat sink, the component is running too hot with the junction temperature exceeded.

Same applies to the tires of your mountain bike and car.

 

[nasu]

No matter at what temperature the water evaporates, there is some vaporization heat (latent heat) used. The vaporization heat varies with temperature. Usually, it is given at the boiling point but this does not mean that at lower temperatures does not exist or there is no energy required for evaporation. Evaporation at room temperature would result in the cooling of the rest of the water if there were no heat transferred continuously from the environment. You can see this effect when the evaporation is too fast for the thermal equilibrium to be achieved, like when you have very volatile liquids like alcohol or acetone. So, no matter at what temperature the evaporation happens you need to account for the vaporization heat. For introductory physics problems, we assume that no evaporation takes place until the boiling point. If you consider a continuous process with evaporation taking place at all temperatures until the boiling point you need to take into account the variation of the vaporization heat with temperature and you need to know the evaporation rate.