What Is the Temperature of Water in Kettle 1?

AI Thread Summary
The discussion revolves around calculating the temperature of water in a kettle, specifically addressing the energy dynamics involved when heating water. Participants analyze the energy required to heat water, factoring in the kettle's heat contribution and efficiency losses. There is a focus on the specific calculations for two scenarios: one with an initial hot kettle and another with cold water added. Participants request clearer explanations and working out of the calculations, emphasizing the importance of understanding energy loss and efficiency. The conversation highlights the need for precise data and methodology in solving the problem effectively.
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1. Please see the attachments for my attempts at answering the question.

forum question.PNG


Please note that this is not a homework question as the answers are provided. I just want to know how to get there. Thanks in advance.
 

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Not sure how you're interpreting the second question. My interpretation is:
  • The 20% of power that was not useful in the first part includes the heat required to raise the kettle to 90C.
  • In the second part, the hot 1 litre of water is poured out and a second litre of water at 15C is poured in.
  • The wasted heat is less the second time because the kettle itself starts hot.
Your attachments are not easy to read. They come out side-on and the writing is a bit small. The ability to add images is really intended for diagrams and extracts from printed matter. Please take the trouble to type in working.
 
I have the same question in my study guide and I get 10 min and 7 seconds for part b
 
DevonZA said:
I have the same question in my study guide and I get 10 min and 7 seconds for part b
I get 11, 58 for part a, but maybe 4.2 J per Calorie is not accurate enough. For part b I get 10, 24.
Please post your working.
 
haruspex said:
I get 11, 58 for part a, but maybe 4.2 J per Calorie is not accurate enough. For part b I get 10, 24.
Please post your working.

1000g-150g = 850g

Eout = mct
= (.85)(4190)(75)
= 267112.5J

Ein = Eout/efficiency
= 267112.5/0.8
= 333890.625Jtime = Ein/Pin
= 333890.625/550
= 607s = 10 min and 7 sec
 
DevonZA said:
1000g-150g = 850g

Eout = mct
= (.85)(4190)(75)
= 267112.5J

Ein = Eout/efficiency
= 267112.5/0.8
= 333890.625Jtime = Ein/Pin
= 333890.625/550
= 607s = 10 min and 7 sec
Why do you subtract 150 from 1000?
 
haruspex said:
Why do you subtract 150 from 1000?

The question states that the water equivalent of the kettle is 150g. 1 litre = 1000g. Therefore the kettle only needs to heat 1000g-150g=850g for the second litre of water. I'm not sure if I'm correct though.
 
DevonZA said:
The question states that the water equivalent of the kettle is 150g. 1 litre = 1000g. Therefore the kettle only needs to heat 1000g-150g=850g for the second litre of water.
There's a litre of water in both cases. The kettle is additional to that.
You need to think about where the other 20% of the power goes. In each case, the equation takes the form:
energy in = energy to heat water + energy to heat kettle + energy lost to room/air
Assign symbols to those (distinguishing a and b) and write the equation for each using the given data.
 
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