Physics Problem - Total Heat Loss in a House w/60 People

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The discussion revolves around calculating the temperature inside a house with 60 people, given its total heat loss of 500 W per degree Celsius and an outside temperature of 10 degrees Celsius. Each person generates 100 W of heat, resulting in a total of 6000 W from the occupants. To determine the indoor temperature, the heat transfer formula Q = U x A x ΔT is suggested, where U is the overall heat transfer coefficient and A is the surface area of the house. The challenge lies in accurately calculating U, which requires understanding the thermal resistance of the house's components. The conversation emphasizes the complexity of the problem due to the non-linear nature of heat loss with temperature differences.
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OK, I have a question that I am trying to do for physics that I just can't figure out. I don't even know how to start it, so any help in getting started would be much appreciated. Here it is:

The total heat loss for a house is 500 W/(Degrees Celsius). The temperature outside the house is 10 (Degrees Celsius). Assume that the average power output for a human body is: 100 W. If there is no heat sources in the house, what will the temperature of the house be if there are 60 people in the house?

If anyone knows how to do the problem, or any ideas of where I should start, that would be great.

Thanks in advance.
 
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JoshHull said:
The total heat loss for a house is 500 W/(Degrees Celsius). The temperature outside the house is 10 (Degrees Celsius). Assume that the average power output for a human body is: 100 W. If there is no heat sources in the house, what will the temperature of the house be if there are 60 people in the house?
This is a confusion statement of the problem. Can you give us the actual wording of the problem?

It appears to be a kind of simplified blackbody radiation problem.

If the total heat loss is 500 Watts/degree C of temperature difference and there are 6000 watts of heat being produced in the house, then the temperature will rise until 6000 watts are being lost. So the temperature will rise until there is a temperature difference of 12 degrees C. with the outside.

The problem is that it is not that simple. The rate of radiation loss increases as the fourth power of the temperature difference.

AM
 


To start this problem, we can use the formula for heat transfer: Q = U x A x ΔT, where Q is the total heat loss, U is the overall heat transfer coefficient, A is the surface area of the house, and ΔT is the temperature difference between inside and outside the house.

In this case, we know that Q = 500 W/(Degrees Celsius), U is unknown, A is the surface area of the house, and ΔT = (Tinside - 10). We also know that there are 60 people in the house, each producing an average power output of 100 W.

To find the overall heat transfer coefficient (U), we can use the formula: U = 1/Rtotal, where Rtotal is the total thermal resistance of the house. The thermal resistance can be calculated by adding the thermal resistance of each component of the house (walls, windows, roof, etc.).

Once we have the value of U, we can plug it into the formula Q = U x A x ΔT to solve for Tinside.

I hope this helps you get started on the problem. Let me know if you need any further clarification. Good luck!
 
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