Heat Energy and Power -- Heating ice in a copper pan

AI Thread Summary
To calculate the power needed to heat a 1360-gram chunk of ice in a 600-gram copper pan from 0°C to 35°C in 10 minutes, the relevant equations include Q=mcΔT for the copper and Q=mLf for the ice. The total heat energy required combines the heat absorbed by the ice as it melts and the heat absorbed by the copper as it warms up. The final state of the ice is water at 35°C, which is essential for determining the total energy needed. The power can then be calculated by dividing the total energy by the time of 10 minutes. Understanding the final state of the ice and the temperature of the entire system is crucial for accurate calculations.
Madelin Pierce
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Homework Statement


Calculate the power needed to heat a 1360 gram chunk of ice in a 600 gram copper pan from 0 ◦C to the system’s final temperature of 35◦C in a time of 10 minutes.

Homework Equations


Q=mcchange in temp
Q=mLf
P= energy/time

The Attempt at a Solution


I thought that adding the heat energy of ice, q=mLf, with the copper's heat, q=mc change in temp, and then dividing by time. Or would subtract copper from ice? And what do I do when it says temp of whole system?
 
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Madelin Pierce said:
thought that adding the heat energy of ice, q=mLf, with the copper's heat, q=mc change in temp, ...
... plus one more thing.
 
haruspex said:
... plus one more thing.
I don't what it is
 
Madelin Pierce said:
I don't what it is
What is the final state of the ice?
 
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haruspex said:
What is the final state of the ice?
water
 
Madelin Pierce said:
water
At what temperature?
 
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