Calculating Water Evaporation Time: Heat Transfer Question Explained

In summary, the conversation discusses how to find the time needed to raise the temperature of 1 litre of water to boiling point using the equation Δu=mCΔT, where Δu is the energy input by the electric burner (500W), m is the mass of water (1 kg), C is the specific heat capacity of water (336 J/kgK), and ΔT is the temperature difference (78K). It is mentioned that the latent heat of vaporisation of water is needed to calculate the time needed for the water to evaporate. The conversation also clarifies that in part one, specific heat capacity is used, while in part two, latent heat of vaporisation is used.
  • #1
CannonSLX
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Homework Statement



1litre of water is heated by an electric burner which couples 500W into the water. How long will it take to raise the temperature of the water to boiling point?

How much longer will it take for the water to evaporate?

Homework Equations


Δu=mCΔT, where

The Attempt at a Solution


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To complete this question, don't I need to know the latent heat capacity of water ?
I understand that I can calculate the time as Δu=mCΔT, where Δu= (500x(t)) J and rearrange to solve for t, but am I missing a variable or is there another way to solve this ?Thanks for your time :)
 
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  • #2
To complete this question, don't I need to know the latent heat capacity of water ?
What has to happen for latent heat to be important?
When does that happen in the question?
Therefore: when do you need the latent heat equation?

I understand that I can calculate the time as Δu=mCΔT, where Δu= (500x(t)) J and rearrange to solve for t, but am I missing a variable or is there another way to solve this ?
Can you identify the missing variable?
 
  • #3
Simon Bridge said:
What has to happen for latent heat to be important?
When does that happen in the question?
Therefore: when do you need the latent heat equation?

Can you identify the missing variable?
So I don't need the latent heat value, and so the equation I have listed is not the one to use ?
 
  • #4
CannonSLX said:
So I don't need the latent heat value, and so the equation I have listed is not the one to use ?
Simon didn't say you don't need it. There are two parts to the question. Can you do the first part, or do you think there is something missing for that, and if so what? At what stage do you think you need the latent heat value?
 
  • #5
haruspex said:
Simon didn't say you don't need it. There are two parts to the question. Can you do the first part, or do you think there is something missing for that, and if so what? At what stage do you think you need the latent heat value?

I'd need the latent heat value to find the how much water will evaporate, so for the first part, would is the missing variable the temperature ?
 
  • #6
CannonSLX said:
I'd need the latent heat value to find the how much water will evaporate, so for the first part, would is the missing variable the temperature ?
Right on both counts. So you cannot proceed with the first part without more information, but you can look up the latent heat of evaporation and do the second part.
 
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  • #7
haruspex said:
Right on both counts. So you cannot proceed with the first part without more information, but you can look up the latent heat of evaporation and do the second part.
How would I work out the temperature difference ?
And could I not then calculate the latent heat of vaporisation once I have the temperature ?
 
  • #8
CannonSLX said:
How would I work out the temperature difference ?
And could I not then calculate the latent heat of vaporisation once I have the temperature ?
I don't understand the first question. As I wrote, you cannot work out the temperature difference if you are not told the start temperature, so you cannot do the first part.
For the second part you do not need to know the starting temperature of the water. You are considering the water having reached boiling point and only need to work out how much energy is required to vaporise it.
The latent heat of vaporisation of water is a constant; look it up.
 
  • #9
haruspex said:
I don't understand the first question. As I wrote, you cannot work out the temperature difference if you are not told the start temperature, so you cannot do the first part.
For the second part you do not need to know the starting temperature of the water. You are considering the water having reached boiling point and only need to work out how much energy is required to vaporise it.
The latent heat of vaporisation of water is a constant; look it up.
Oh I see. Apologies for the misunderstanding.
Thanks for explaining :)
 
  • #10
When not given data, it is usually reasonable to assume an obvious choice (clearly stated). In this case, room temperature (say, 22degC) is a defensible choice.
 
  • #11
insightful said:
When not given data, it is usually reasonable to assume an obvious choice (clearly stated). In this case, room temperature (say, 22degC) is a defensible choice.
So to workout the time needed to raise the temperature to boiling (100 Deg'C) if I assume initial temp is 22 Deg'C, ΔT = 78K.
And I know the power input is (500Wxt). Is there a particular equation I can use to solve ?
 
  • #12
CannonSLX said:
Is there a particular equation I can use to solve ?
Your Relevant equation in your first post is all you need for this part. Be careful with unit conversions. Check and double-check each step.
 
  • #13
insightful said:
Your Relevant equation in your first post is all you need for this part. Be careful with unit conversions. Check and double-check each step.

So Δu=mCΔT and given C=336 [Latent heat of fusion] (apologies for no units, using my phone makes it tedious to type)
(500w x t) = 1 x 336 x 78 ? and then I just solve for t ?

Then to find how much long it will take to evaporate, would I use the Latent heat of vaporisation value in the same equation ?
 
  • #14
CannonSLX said:
So Δu=mCΔT and given C=336 [Latent heat of fusion]
"Fusion" means "melting." C is heat capacity. Please be more diligent in these critical, basic details.
 
  • #15
insightful said:
"Fusion" means "melting." C is heat capacity. Please be more diligent in these critical, basic details.
So part one of the question uses specific heat capacity and part two uses latent heat of vaporisation ?
 
  • #16
So part one of the question uses specific heat capacity and part two uses latent heat of vaporisation ?
... this is something you can answer yourself by looking up the definitions of these terms.

Specific heat capacity is the amount of work heat needed to change the temperature of 1 unit mass of stuff by one unit of temperature.
Latent heat is the amount of heat needed to change the state of 1 unit mass of stuff without a change in temperature.
Now you tell me: which one do you need for the equation where the state does not change but the temperature does?

If you didn't know these definitions already - then you have a serious problem: go back over your text and course notes and revise.
 
  • #17
Simon Bridge said:
... this is something you can answer yourself by looking up the definitions of these terms.

Specific heat capacity is the amount of work heat needed to change the temperature of 1 unit mass of stuff by one unit of temperature.
Latent heat is the amount of heat needed to change the state of 1 unit mass of stuff without a change in temperature.
Now you tell me: which one do you need for the equation where the state does not change but the temperature does?

If you didn't know these definitions already - then you have a serious problem: go back over your text and course notes and revise.
ΔQ=mL ?
 
  • #18
CannonSLX said:
ΔQ=mL ?
Assuming L is the latent heat of vaporization, yes (for the second part of the problem).
 
  • #19
insightful said:
Assuming L is the latent heat of vaporization, yes (for the second part of the problem).
So ΔQ= 500w x t = mL and just solve for t ?
 
  • #20
CannonSLX said:
So ΔQ= 500w x t = mL and just solve for t ?
Yes.
 
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  • #21
insightful said:
Yes.
Thank you. Got there eventually :)
 
  • #22
Well done.
 

Related to Calculating Water Evaporation Time: Heat Transfer Question Explained

1. How is water evaporation time calculated?

The water evaporation time can be calculated by using the following formula: Time = (Mass of Water x Specific Heat x Temperature Difference) / (Heat Transfer Rate x Surface Area). This formula takes into account the mass of water, its specific heat, the temperature difference between the water and its surroundings, the heat transfer rate, and the surface area of the water.

2. What factors affect the water evaporation time?

The water evaporation time is affected by various factors such as temperature, humidity, surface area, and air flow. Higher temperatures and lower humidity levels will result in faster evaporation, while a larger surface area and increased air flow will also speed up the process.

3. How does heat transfer impact water evaporation time?

Heat transfer is a crucial factor in determining the water evaporation time. The rate at which heat is transferred from the surroundings to the water will affect the temperature difference between the two, which in turn affects the evaporation rate. Higher heat transfer rates will result in faster evaporation.

4. Can the water evaporation time be accurately predicted?

While the formula for calculating water evaporation time provides a good estimate, there are many variables that can affect the actual evaporation time. These include changes in temperature, humidity, and air flow, as well as the presence of impurities in the water. Therefore, the predicted evaporation time may not always be completely accurate.

5. How can the water evaporation time be shortened?

To shorten the water evaporation time, one can increase the temperature, decrease the humidity, increase the surface area, or increase the air flow. Additionally, using a fan or other means of air circulation can also help to speed up the process. However, it is important to note that evaporation is a natural process and cannot be completely controlled or manipulated.

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