Heat Loss

  • Thread starter Cheddar
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  • #1
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Homework Statement


A metal (doesn't say what metal) is in a heater that is set to 700 K.
If the temperature of the ball is 900 K then the rate of heat loss = 0.10 J/min.
What is the rate of heat loss when the ball's temperature = 800 K.
The ball's emissivity doesn't change appreciably with temperature.


Homework Equations


Q = emissivity * Boltzman Law * Temperature (to the 4th power in K) * Area * time


The Attempt at a Solution


I solved for A using the given information then plugged it in to the equation above to solve for Q/t using 800 K instead of the 900 K. My answer doesn't check out. Am I suppose to convert 0.10 J/min into J/seconds? If so, how?
 

Answers and Replies

  • #2
tiny-tim
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Hi Cheddar! :smile:
… If the temperature of the ball is 900 K then the rate of heat loss = 0.10 J/min.

My answer doesn't check out. Am I suppose to convert 0.10 J/min into J/seconds? If so, how?

Yes, all your measurements must be in SI units (kg, second, joule, etc).

Calculations can go horribly wrong if you don't do that!

To convert J/min into J/second, just do it the same way you'd convert feet/min into feet/second. :wink:
 
  • #3
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Thank you. I did that and the answer still doesn't match up though.
I get an area of 4.48018196 * 10(-8 power). Plugging this into:
Q/t = emissivity * Boltzman's * Temperature(4th power) * Area
leaves me with 0.0010404918, which doesn't check out.
 
  • #4
Redbelly98
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Homework Equations


Q = emissivity * Boltzman Law * Temperature (to the 4th power in K) * Area * time

This equation isn't quite right. Question: what would the rate of heat loss be if the ball were at the same temperature as the heater in which it is located?
 
  • #5
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So, instead of T(4th power) it should maybe be the difference between the ball's temp and the environments temp(to the 4th power)??
 
  • #6
Redbelly98
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Close. It would be (Tball4 - Tenvir4)
 
  • #7
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I think I have it.
0.40 J/min
 
  • #8
Redbelly98
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The heat loss rate should be less at a lower temperature.
 

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