Calculating Rate of Heat Loss for Metal at 800K

[Cheddar]

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 into 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?

 

[tiny-tim]

Hi Cheddar!

… 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.

 

[Cheddar]

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.

 

[Redbelly98]

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?

 

[Cheddar]

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)??

 

[Redbelly98]

Close. It would be (T_ball⁴ - T_envir⁴)

 

[Cheddar]

I think I have it.
0.40 J/min

 

[Redbelly98]

The heat loss rate should be less at a lower temperature.