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Thermodynamics, calculating the time it will take water to boil away

by madjockmcfers
Tags: boil, thermodynamics, time, water
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madjockmcfers
#1
Oct11-11, 04:10 PM
P: 3
1. The problem statement, all variables and given/known data

Ok, so my book has the following problem: A kettle can heat 20C water to 100C in 5 minutes. How long will it take to completely boil it away assuming the same rate of heat addition.

2. Relevant equations

I think Q=mcΔT is relevant but beyond that I am struggling.

3. The attempt at a solution

Hi, I've just enrolled in my first ever physics class at the age of 34. I've been doing well but this has stumped me, any guidance as to how to even start the question would be very useful. I feel that once I get a start I can finish it myself.

So, I've worked through the equation using Q= 2260kJ as this is the heat vaporisation of water, c i've assumed to be 4186J (the specific heat of water) and Δ T is 80C. This gives me a mass of 6.75kg.

The rate of heat addition is 80C / 5 minutes = 16C per minute.

However, I'm not sure I've used the correct values, and I'm not certain on how to proceed from here. I'm not even sure I need to know the mass. The problem is the book (Energy, its use and the Environment) doesn't have any examples like this, and I am completely new to physics.

I've been doing well so far but this has completely thrown me.

Perhaps I should use the formulae W(work)=Qh-Qc where Qh equals end temperature and Qc equals initial temperature...and work out from that the amount of work done on the water and then somehow get to how much work needs to be done to boil away the water....

But without being given the mass how do I proceed? Ugh....I am doing well in this course but this has stumped me.
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Spinnor
#2
Oct11-11, 08:53 PM
P: 1,368
The quantity of water is unknown, so just assume it is 1 kg to make the calculation simple.

ΔQ = [4.18kJ/kgC]*1kg*80degreesC = 334.4kJ

Power = ΔQ/Δt = 334.4kJ/300s = 1.11kJ/s = 1110 watts = 1.11kw

My book gives the heat of vaporization of water as 2256 J/g or 2256kJ/kg.

So 1.11kw * Δt = 2256kJ

So Δt = 2030 s = about 34 minutes ?

That seems long???
madjockmcfers
#3
Oct11-11, 09:39 PM
P: 3
It does seem a little long.

One problem is clearly it doesn't give a mass.....so how I'm supposed to give a correct answer I don't know. My only thought is that mass cancels out somewhere...I just haven't worked out where yet.

The additional problem here is the question says how long will it take to boil at the same rate of temperature increase....which is (100C-20C) / 5 mins = 16C increase per minute. So, it'll take less than 34 minutes because the temperature of the steam is increasing by 16C per minute.

JustinLiang
#4
Oct11-11, 10:06 PM
P: 82
Thermodynamics, calculating the time it will take water to boil away

Quote Quote by Spinnor View Post
The quantity of water is unknown, so just assume it is 1 kg to make the calculation simple.

ΔQ = [4.18kJ/kgC]*1kg*80degreesC = 334.4kJ

Power = ΔQ/Δt = 334.4kJ/300s = 1.11kJ/s = 1110 watts = 1.11kw

My book gives the heat of vaporization of water as 2256 J/g or 2256kJ/kg.

So 1.11kw * Δt = 2256kJ

So Δt = 2030 s = about 34 minutes ?

That seems long???
I got 2019.1s, I am not 100% sure I did it right as well since I am still learning this but...

I first found the power with an unknown variable m.
Q=mcΔT
Q=m(4190)(80)
P=m(4190)(80)/300s

Now because we know the energy to vaporize all the water is Q=mL. You take that number and divide it by the power, thus cancelling the mass. What do you guys think? Makes sense to me. Vaporizing water would require more energy than heating it up, thus the answer must be greater than 5 minutes. In this case about 34 minutes.
gneill
#5
Oct11-11, 10:12 PM
Mentor
P: 11,689
Don't forget to include the time it takes to heat the water from 20C to 100C.
madjockmcfers
#6
Oct12-11, 02:44 PM
P: 3
Thanks everyone. That makes sense. I was struggling yesterday but in the cold light of day that looks about right!


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