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Heat transfer help.

  • Thread starter -Dragoon-
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  • #1
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Homework Statement


A 0.25 kg piece of ice is warmed by an electric heater. Assume that there has been no loss of energy to the surroundings. It takes 150 seconds to heat the ice from -30° to -10°. How much additional time after 150 seconds will be required to melt all of the ice, assuming that the power of the heater is constant?

I made this on my computer to help me visualize the problem:
http://img824.imageshack.us/i/energytransferphysics.jpg/

Homework Equations


Qwarm ice = miceciceΔtice
Q melt ice = miceLfusion
P = ∆Q/∆t

The Attempt at a Solution


Firstly, I use the first equation to find the energy it takes to heat the ice to the melting -30° to -10° which yields about 10500 J. I know this takes a 150 seconds, so I use the third equation to find the power which gives a value of 70 W. Now that I have the power, I can find the time it takes to heat the ice from -10° to 0° using the first equation. I use the first equation again to get a value of 5250 J and divide this by 70 W, which is 75 seconds to heat from -10° to 0°. Finally, I use the second equation to find the energy needed to melt the ice and gives me a value of 82,500 J. I divide this by the power (70W) to find the energy and finally yield a value of 1100 S. 1100+75 = 1175 seconds. After the 150 seconds it took to heat the ice from -30° to -10°, it will take another 1175 seconds for the ice to heat from -10° to 0° and melt.

Did I do this correctly?
 

Answers and Replies

  • #2
collinsmark
Homework Helper
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Firstly, I use the first equation to find the energy it takes to heat the ice to the melting -30° to -10° which yields about 10500 J. I know this takes a 150 seconds, so I use the third equation to find the power which gives a value of 70 W. Now that I have the power, I can find the time it takes to heat the ice from -10° to 0° using the first equation. I use the first equation again to get a value of 5250 J and divide this by 70 W, which is 75 seconds to heat from -10° to 0°. Finally, I use the second equation to find the energy needed to melt the ice and gives me a value of 82,500 J. I divide this by the power (70W) to find the energy and finally yield a value of 1100 S. 1100+75 = 1175 seconds. After the 150 seconds it took to heat the ice from -30° to -10°, it will take another 1175 seconds for the ice to heat from -10° to 0° and melt.

Did I do this correctly?
Your general approach seems perfectly fine to me. :approve:

But I think you're making some significant rounding/truncation errors. For example, you've implied that
82,500/70 → 1100.​
That's significantly different that what I would say, which would be closer to something around 1179.

Anyway, just be careful with your significant figures. :wink:
 
  • #3
309
7
Your general approach seems perfectly fine to me. :approve:

But I think you're making some significant rounding/truncation errors. For example, you've implied that
82,500/70 → 1100.​
That's significantly different that what I would say, which would be closer to something around 1179.

Anyway, just be careful with your significant figures. :wink:
Ah, I see where I made the mistake.

Thanks for the help! :biggrin:
 

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