The discussion revolves around a heat transfer problem involving a cooler at a beach, focusing on the calculation of heat transfer through the cooler's insulation. The participants explore the application of the heat conduction equation and the implications of various parameters such as surface area, temperature difference, and insulation thickness.
There is an ongoing exploration of the heat transfer equations, with participants providing insights and corrections regarding the parameters involved. Some participants have suggested alternative equations and clarified the definitions of terms, while others are verifying their calculations and assumptions. Multiple interpretations of the problem's dimensions are being considered.
Participants note potential ambiguities in the problem statement, particularly regarding whether the dimensions provided refer to the inside or outside of the cooler. This uncertainty influences their calculations and assumptions about the heat transfer process.
Hey, that L is not latent heat in equation but length.goonking said:
oh, sorry! good catchRaghav Gupta said:
There is another equation for heat transfer what is that?goonking said:
Q = L mRaghav Gupta said:
How to bring time term here which is 3 hrs?goonking said:
i think the answer we get, if we did all the algebra, is the amount of heat per seconds.Raghav Gupta said:
Q = (0.035) (0.02m x 0.4m) (10800 seconds) (29 C) / (0.4m) = 219.24 joulesRaghav Gupta said:
Yes, it is correct.goonking said:
goonking said:
oh, i forgot to multiply the area by 6, so it should be 6( 0.02m x 0.4m) = ACWatters said:
goonking said:
ok, got the answer to be 1.57 kg, sounds much more reasonable nowCWatters said:
should be 0.02 mCWatters said:
make sure you put in 0.02 m, not 0.2mRaghav Gupta said:
true, i was thinking that too when I tried to draw out the coolerCWatters said:
Isn't the answer given in your problem so you could verify?goonking said:
yes, the answer was 1.57 kgRaghav Gupta said:
