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# Heat capacity question - where have I gone wrong? watch

1. Hi guys,

The question says: An electric kettle has a heat capacity of 450JK^-1 and an element whose power is 2.25kW. Ignoring losses of energy to the surroundings what is the rate of rise of temperature when the kettle contains 1kg of water?

The answer is 29Kmin^-1 but I got 32Kmin^-1 I don't know whether the answer is wrong or I'm wrong, could someone else please tell me what they get?

Thanks
2. (Original post by Blake Jones)
Hi guys,

The question says: An electric kettle has a heat capacity of 450JK^-1 and an element whose power is 2.25kW. Ignoring losses of energy to the surroundings what is the rate of rise of temperature when the kettle contains 1kg of water?

The answer is 29Kmin^-1 but I got 32Kmin^-1 I don't know whether the answer is wrong or I'm wrong, could someone else please tell me what they get?

Thanks
How did you get 32 K min^-1z? I can get 29 K min^-1...
3. (Original post by Eimmanuel)
How did you get 32 K min^-1z? I can get 29 K min^-1...
I did this...

P=2.25kW = 2250Js^-1
Change in Q = Pt
Change in Q = mc change in temp
c = 4200
m= 1
Change in Q = 4200 change in temp
Change in Q / t = P
If t = 1
2250/4200 = change in temp = 0.536Ks^-1
0.536 * 60 = 32Kmin^-1
4. (Original post by Blake Jones)
I did this...

P=2.25kW = 2250Js^-1
Change in Q = Pt
Change in Q = mc change in temp
c = 4200
m= 1
Change in Q = 4200 change in temp
Change in Q / t = P
If t = 1
2250/4200 = change in temp = 0.536Ks^-1
0.536 * 60 = 32Kmin^-1
You need to include the thermal energy gained by the kettle.
5. (Original post by Eimmanuel)
You need to include the thermal energy gained by the kettle.
Oh okay, how should I do this?
6. (Original post by Blake Jones)
Oh okay, how should I do this?
Thermal energy gained by the kettle = heat capacity of kettle * change in temp.
7. (Original post by Eimmanuel)
Thermal energy gained by the kettle = heat capacity of kettle * change in temp.
Ah, awesome, very helpful, thank you!
8. (Original post by Eimmanuel)
How did you get 32 K min^-1z? I can get 29 K min^-1...
How did you find the change in temperature of the kettle for the Thermal energy gained by the kettle part?
9. (Original post by Blake Jones)
How did you find the change in temperature of the kettle for the Thermal energy gained by the kettle part?
Use conservation of energy,

P * 60 s = mass of water * specific heat capacity * delta T + heat capacity of kettle * delta T

The change in temperature (delta T) is within 1 min.

Hope it is clear.
10. (Original post by Eimmanuel)
Use conservation of energy,

P * 60 s = mass of water * specific heat capacity * delta T + heat capacity of kettle * delta T

The change in temperature (delta T) is within 1 min.

Hope it is clear.
What units are you using for the heat capacity of the kettle?
I still can't get it right.
I did 2.25kW = 2250W
2250*60=135000
135000 = 4200 change in temp + 0.45*change in temp
which gives me a change in temp of 22
11. (Original post by Blake Jones)
What units are you using for the heat capacity of the kettle?
I still can't get it right.
I did 2.25kW = 2250W
2250*60=135000
135000 = 4200 change in temp + 0.45*change in temp
which gives me a change in temp of 22

The question said "An electric kettle has a heat capacity of ...."

Why do you divide 450 by 1000?
12. (Original post by Eimmanuel)

The question said "An electric kettle has a heat capacity of ...."

Why do you divide 450 by 1000?

So 135000/4200 = 451T
32...../451 = 0.07 = change in T
13. (Original post by Blake Jones)

So 135000/4200 = 451T
32...../451 = 0.07 = change in T
I hope this is not your final working to

If yes, below is the final answer.
Spoiler:
Show

14. (Original post by Eimmanuel)
I hope this is not your final working to

If yes, below is the final answer.
Spoiler:
Show

Oh! I see! Thank you so much! There is no way I would have managed that question without you, I've got to give rep to someone else first apparently but I'll definitely rep you soon!
15. in 60 seconds the electric energy delivered to the kettle & water is*

60 * 2250 J

the water needs 4184 J to rise one degree C

the kettle needs 450 J to rise one degree C

thus the kettle & water together need 4634 J to rise one degree C

so { 60 * 2250 } / 4634 gives the number of degrees C the kettle & its contents rise in one minute

*
*
16. (Original post by the bear)
in 60 seconds the electric energy delivered to the kettle & water is*

60 * 2250 J

the water needs 4184 J to rise one degree C

the kettle needs 450 J to rise one degree C

thus the kettle & water together need 4634 J to rise one degree C

so { 60 * 2250 } / 4634 gives the number of degrees C the kettle & its contents rise in one minute

*
*
Thank you, I feel so stupid
17. (Original post by Blake Jones)
Thank you, I feel so stupid
it is how we all learn... *
18. (Original post by the bear)
it is how we all learn... *
That's very true

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