Please confirm answer to thermodynamic question

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The discussion revolves around a thermodynamic problem involving heat loss through a glass roof of a greenhouse. Given the heat loss rate of 120 kW and the internal temperature of 25 degrees Celsius, the calculation suggests an outside temperature of 55 degrees Celsius, which is questioned by the poster. Participants emphasize understanding the direction of heat flow, noting that heat moves from warmer areas to cooler ones, indicating that the greenhouse is designed to maintain a higher internal temperature. The importance of correctly interpreting the problem's context is highlighted, urging a reevaluation of the calculations based on the greenhouse's purpose. The conversation underscores the need for clarity in thermodynamic principles related to heat transfer.
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Homework Statement



The rate of heat loss to the outside through a 10mm-thick glass roof of a green house is 120kW. The length and breadth of the glass material of the roof are 20m and 2m, respectively. Calculate the temperature of the outside if the temperature inside the green-house is 25 degrees C.

Assume thermal conductivity of the glass material is 1.0W/m/degrees C.

Homework Equations



dQ/dt = kA(dT/dx)

The Attempt at a Solution



dQ/dt = 120000W, k = 1W/m/degree C, A = 20mx2m=40m^2 and dT/dx = (T-25degree C)/10mm = (T-25)/0.01m

T = 55 degree CI'm not sure this is right, any help is much appreciated!

Thanks in advance!
 
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Where is warmer? Inside or outside of the greenhouse?

ehild
 
It doesn't state!

The exact question is written above.

Thanks!
 
What do you think? What is a greenhouse for?

It was said that the heat loss of the greenhouse is 120 kW. The greenhouse loses energy, heat flows out of it. What is the direction of heat flow? From a warmer place to a colder one or from a cold place to a warm one?


ehild
 
Heat loss:

\frac{dQ}{dt} < 0

Heat gain:

\frac{dQ}{dt} > 0

You must interpret the question and understand that a greenhouse is a structure to regulate and maintain a certain temperature in the inside, usually hotter than the outside environment.

Do your calculations again based on that.
 
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