Heating water in a kettle (time, boil out time, work done)

[moenste]

Homework Statement

A domestic kettle is marked 250 V, 2.3 kW and the manufacturer claims that it will heat a pint of water to boiling point in 94 s.

(a) Test this claim by calculation and state any simplifying assumptions you make.
(b) If the kettle is left switched on after it boils, how long will it take to boil away half a pint of water measured from when it first boils?
(c) Estimate the work done against an atmospheric pressure of 100 kPa when 1 cm³ of water evaporates at 100 °C, producing 1600 cm³ of steam. Express this as a percentage of the total energy required to evaporate 1 cm³ of water at 100 °C.

• Specific heat capacity of water = 4.2 * 10³ J kg^-1 K^-1
• Specific latent heat of vaporisation of water = 2.3 * 10⁶ J kg^-1
• Density of water = 1.0 g cm^-3
• 1 pint = 570 cm³

Answers: (c) 1.6 * 10² J, 7.0 %

2. The attempt at a solution
(a) We need to find Δθ in ΔQ = m c Δθ. Q is V I t, where I = P / V = 2300 / 250 = 9.2 A.

Then find Q = 250 * 9.2 * 94 = 216 200 J. m is 1 g cm^-3 * 570 cm³ = 570 g. 570 g / 1000 = 0.57 kg. Substitute for Δθ = ΔQ / mc = 216 200 / (0.57 * 4.2 * 10³) = 90.3 °C.

Water boils at 100 °C, so the manufacturer is wrong. In 94 s we'll only get to 90.3 °C.

(b) The mass of water that will be in the kettle at 100 °C: m = ΔQ / cΔθ = 216 200 / (100 * 4.2 * 10³) = 0.51 kg.

Find ΔQ with half of that mass: ΔQ = 0.26 * 4.2 * 10³ * 100 = 108 100 J.

Find time: t = Q / VI = 108 100 / 250 * 9.2 = 47 seconds. That is half the claimed boiling time. Maybe it was possible to just divide the boiling time (94 s) by two? Since at boiling time half of the water will boil out in half the time (same as a third in a third and 1/4 in 1/4 etc.).

I didn't start (c) since I am not sure whether both (a) and (b) are correct. I also worked on this problem some while ago and in (a) I had a different mass = 0.473 kg and I got the temperature to be equal to 109.35 °C (so the manufacturer is right). But I can't remember at all how did I derive that number, maybe it is wrong and this is correct: m is 1 g cm^-3 * 570 cm³ = 570 g. 570 g / 1000 = 0.57 kg.

Any thoughts on (a) and (b) please?

 

[kuruman]

(a) looks correct although you could have gotten the heat going in by using Δθ = P t directly. The problem is asking you to list your assumptions. What assumptions are you using to get your answer?
(b) I would find the mass of the water differently. What is the mass of one pint of water? Use that instead.

 

[moenste]

(a) looks correct although you could have gotten the heat going in by using Δθ = P t directly. The problem is asking you to list your assumptions. What assumptions are you using to get your answer?
(b) I would find the mass of the water differently. What is the mass of one pint of water? Use that instead.

You meant ΔQ and not Δθ? Since P t is 216 200, same as Q. In terms of assumptions I didn't make any. Water boils at 100 degrees, so if the manufacturer is right then the temperature in 94 seconds will be the same or higher than 100. If wrong -- less.

1 pint = 570 cm³
density of water = 1.0 g cm^-3

570 * 1 = 570 g. In kg: 570 / 1000 = 0.57 kg.

 

[kuruman]

You meant ΔQ and not Δθ?

Yes, I meant ΔQ, sorry for the confusion.

1 pint = 570 cm³

Where did you get this number? Google "pint to cc conversion" and see what you get.

On edit: Which side of the Atlantic are you on? One UK pint is 570 cc as you said, but 1 US pint is 470 cc. More importantly, which side of the Atlantic is the manufacturer of the heater who made this claim? Here is an assumption to add to the list.

 

[moenste]

Where did you get this number? Google "pint to cc conversion" and see what you get.

On edit: Which side of the Atlantic are you on? One UK pint is 570 cc as you said, but 1 US pint is 470 cc. More importantly, which side of the Atlantic is the manufacturer of the heater who made this claim? Here is an assumption to add to the list.

This number is given in the problem (along with others), so I did not look it up.

• Specific heat capacity of water = 4.2 * 103 J kg-1 K-1
  
• Specific latent heat of vaporisation of water = 2.3 * 106 J kg-1
• Density of water = 1.0 g cm-3
• 1 pint = 570 cm3

The book is published in the UK, so I guess that's the reason for 570 cm³. Personally I am from Russia and we don't use pints, litres like the rest of Continental Europe.

1 US pint is 470 cc

I think that that's the reason why in my previous tries I had the mass as 0.473 kg. I probably googled what one pint is and got 470 cm³. And using this number the temperature is 109 °C and the manufacturer is then correct. So we could assume that if the manufacturer is from the US, then the timing is true. And if the manufacturer is from the UK, then it's is wrong. If not your tip I would've never though of that actually, since a person from UK or USA is required to know these differences :), or at least someone who is familiar with the UK / USA differences.

 

[kuruman]

OK, so one assumption is that the amount of water is 1 UK pint which has a mass of 0.57 kg. That's what you say in part (a). However, in part (b) you say that the mass is 0.510 kg. You need to fix that to be consistent. Also, it looks like you misunderstood what (b) is asking you to find. The 94 s of part (a) is irrelevant here. Here is the picture: You have 1 pint of boiling water. As soon as the water starts boiling, you start a timer. What does the timer read when half a pint of water has boiled away?

Can you think of the assumptions that go into calculations of this kind other than the US/UK pint difference?

 

[gneill]

I don't see where you've made any assumption about the starting temperature of the water. You've calculated a Δθ, but ffrom what initial temperature is the 'Δ'? Does it matter for your answer to (a)?

 

[moenste]

OK, so one assumption is that the amount of water is 1 UK pint which has a mass of 0.57 kg. That's what you say in part (a). However, in part (b) you say that the mass is 0.510 kg. You need to fix that to be consistent. Also, it looks like you misunderstood what (b) is asking you to find. The 94 s of part (a) is irrelevant here. Here is the picture: You have 1 pint of boiling water. As soon as the water starts boiling, you start a timer. What does the timer read when half a pint of water has boiled away?

Can you think of the assumptions that go into calculations of this kind other than the US/UK pint difference?

I don't see where you've made any assumption about the starting temperature of the water. You've calculated a Δθ, but ffrom what initial temperature is the 'Δ'? Does it matter for your answer to (a)?

Regarding (a):
We're assuming that we are in the UK, since we have 1 pint = 570 cm³. We need to find to what temperature will the kettle boil the water in 94 seconds. Let's assume that the initial temperature is 0 °C.

ΔQ = mcΔθ, where Δθ = ?
To find ΔQ we use: ΔQ = Pt = 2300 * 94 = 216 200 J.
Mass of the water is: 570 cm³ * water density 1 g cm^-3 = 570 g or 0.57 kg.
So Δθ = 216 200 / (0.57 * 4.2 * 10³) = 90.3 °C

The boiling temperature of water is 100 °C and therefore it is either a UK manufacturer is lying or the manufacturer is from the US and the situation in the US is different.

I think this should be it in (a)?

Update: regarding (b): ΔQ = 0.285 [half 1 pint = 0.57 kg] * 4.2 * 10³ * 100 [boiling temp.] = 119 700 J. t = Q / t = 119 700 / 2300 = 52 s. So it will take 52 seconds to boil half the water.

 

[gneill]

The boiling temperature of water is 100 °C and therefore it is either a UK manufacturer is lying or the manufacturer is from the US and the situation in the US is different.

Or the manufacturer will assume that the water starts at room temperature so that his claim looks better and he can claim it's a more typical situation...

edit: Did a quick google search on "average tap water temperature uk" and found out that it is about 7C, but varies with the season.

 

[kuruman]

The boiling temperature of water is 100 °C and therefore it is either a UK manufacturer is lying or the manufacturer is from the US and the situation in the US is different.

Here some more questions to consider
Is the boiling temperature of water 100 ^oC everywhere on the surface of the Earth?
Does all the power from the heater go into boiling water?

 

[moenste]

Or the manufacturer will assume that the water starts at room temperature so that his claim looks better and he can claim it's a more typical situation...

edit: Did a quick google search on "average tap water temperature uk" and found out that it is about 7C, but varies with the season.

That's also a point :).

Here some more questions to consider
Is the boiling temperature of water 100 ^oC everywhere on the surface of the Earth?
Does all the power from the heater go into boiling water?

Yes, it also depends on the height.
Probably part of it is lost somewhere in powering the buttons etc. So we consider that 100 % of power is used to heat water.

What about (b)? ΔQ = 0.285 [half 1 pint = 0.57 kg] * 4.2 * 10³ * 100 [boiling temp.] = 119 700 J. t = Q / t = 119 700 / 2300 = 52 s. So it will take 52 seconds to boil half the water. Is this correct?

 

[haruspex]

What about (b)? ΔQ = 0.285 [half 1 pint = 0.57 kg] * 4.2 * 103 * 100 [boiling temp.] = 119 700 J. t = Q / t = 119 700 / 2300 = 52 s. So it will take 52 seconds to boil half the water. Is this correct

I think you have misread the question. For this part, assume the water is already at boiling point. How long does it take for half of the water to boil away into vapour?

 

[moenste]

I think you have misread the question. For this part, assume the water is already at boiling point. How long does it take for half of the water to boil away into vapour?

Hm, should it be then: ΔQ = ml = (0.57 / 2) * 2.3 * 10⁶ = 655 500 J → t = Q / VI = 655 500 / (250 * 9.2) = 285 s? Isn't it too long?

 

[kuruman]

I don't think so. It's less than 5 minutes.

 

[moenste]

I don't think so. It's less than 5 minutes.

Why 5 minutes?

 

[moenste]

I don't think so. It's less than 5 minutes.

Regarding (c) I think it should be: Work done W = pV (pressure * volume). p = 100 000 Pa and V = 1600 cm³ or 1.6 * 10^-3 m³. So W = 100 000 * 1.6 * 10^-3 = 160 J. This should be correct.

"Express this as a percentage of the total energy required to evaporate 1 cm3 of water at 100 °C." for this one we have ΔQ = ml where m = 1 cm³ * 1 g cm^-3 = 1 g or 1 * 10^-3 kg. So Q = 10^-3 * 2.3 * 10⁶ = 2300 J. (160 / 2300) * 100 = 7 %.

 

[haruspex]

Hm, should it be then: ΔQ = ml = (0.57 / 2) * 2.3 * 10⁶ = 655 500 J → t = Q / VI = 655 500 / (250 * 9.2) = 285 s? Isn't it too long?

Do you cook? Have you ever had to boil a sauce to reduce its volume? Seems to take forever.

 

[moenste]

Do you cook? Have you ever had to boil a sauce to reduce its volume? Seems to take forever.

That's indeed so :).

Alright, so (a) and (b) should be good. And (c)? I think it should be correct.

Regarding (c) I think it should be: Work done W = pV (pressure * volume). p = 100 000 Pa and V = 1600 cm³ or 1.6 * 10^-3 m³. So W = 100 000 * 1.6 * 10^-3 = 160 J. This should be correct.

"Express this as a percentage of the total energy required to evaporate 1 cm3 of water at 100 °C." for this one we have ΔQ = ml where m = 1 cm³ * 1 g cm^-3 = 1 g or 1 * 10^-3 kg. So Q = 10^-3 * 2.3 * 10⁶ = 2300 J. (160 / 2300) * 100 = 7 %.

 

[kuruman]

Why 5 minutes?

Five minutes is 300 s. Your answer is 285 s, less than 5 min.

And (c)? I think it should be correct.

It looks OK.