Question on Specific heat capacity

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blooberrieex
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#1
Report Thread starter 1 year ago
#1
I need help on this question:
A 2kW electric heater supplies energy to a 0.5 kg copper kettle (380J/Kg*per degree celsius *) containing 1kg of water (4200J/Kg*per degree celsius *)
-Calculate the time taken to raise the temperature by 10 degrees and what have you assumed in doing this calculation.
I have tried plugging in the energy transferred of copper and the water added together into the Power equation but gave me a small decimal...
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ak1237
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#2
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Nothing wrong with a decimal. I assume you did the following:

Copper Kettle: 380 * 0.5 * 10 = 1,900 J
Water: 4200 * 1 * 10 = 42,000 J
Total = 43,900 J

Then t = E/p = 43900 / 2000 = 21.95 seconds

This assumes no heat energy is wasted/lost to surroundings.
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