1
$\begingroup$

I got in an argument with my roommate about whether the toaster oven or the regular oven was more efficient, and I attempted to model it assuming a consistent heat transfer rate over the area of the toaster oven, but I wasn't getting results as I would expect(too steep or too shallow of a heat up, also too much of a gain loss). What is the correct way to calculate the heat loss from a box, assuming that there is a heat source producing so many joules of heat every $dt$? Also each face of the box is made of a different material, like glass or steel or nothing(the bottom of the toaster oven, but there's a counter top).

$\endgroup$
  • $\begingroup$ Heat transfer between an object and the surroundings is usually quantified in terms of a heat transfer coefficient at the surface which, when multiplied by the difference in temperature between the surface and surroundings determines the heat flux. The heat transfer coefficient is determined by considering the forced- and natural convection temperature- and velocity variations within the surrounding air. $\endgroup$ – Chet Miller Sep 29 '18 at 0:59
  • $\begingroup$ @HSchmale an easy way to measure this is to set them both to the same temperature and monitor power consumption. Once at set temperature, any additional power use is exactly the amount of heat that's lost to the environment. $\endgroup$ – Al Nejati Sep 29 '18 at 1:08
  • 1
    $\begingroup$ The question could be improved by specifying whether you are only considering the time when each oven is at temperature, versus the entire cycle of heating up / at temperature / cooling down. $\endgroup$ – DrSheldon Sep 29 '18 at 1:26
  • $\begingroup$ @ChesterMiller That sounds like it should be an answer. $\endgroup$ – David Z Sep 29 '18 at 3:40
  • $\begingroup$ @AlNejati That also sounds like it should be an answer. $\endgroup$ – David Z Sep 29 '18 at 3:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.