# Heat Loss

## 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

tiny-tim
Science Advisor
Homework Helper
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. 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
Staff Emeritus
Science Advisor
Homework Helper

## 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?

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
Staff Emeritus
Science Advisor
Homework Helper
Close. It would be (Tball4 - Tenvir4)

I think I have it.
0.40 J/min

Redbelly98
Staff Emeritus
Science Advisor
Homework Helper
The heat loss rate should be less at a lower temperature.