How Can You Calculate Heat Transfer and Energy Costs in Various Scenarios?

AI Thread Summary
The discussion focuses on calculating heat transfer and energy costs in various scenarios, highlighting specific homework problems. For the electric oven, the correct equation for energy calculation is Q = (KAΔT)t/L, leading to an energy use of approximately 2,104,769.76 J and a cost of about $0.05. The second problem involves calculating the temperature difference between blood capillaries and skin using Q = (KAΔT)/L, yielding a ΔT of about 20.83 °C. The third problem requires finding the heat loss ratio between wool and goose down using similar equations, while the fourth problem addresses the temperature of a car parked in the sun, needing clarification on the use of the radiation equation Q/tA = e(5.67x10^-8)(T^4). Overall, the thread emphasizes the importance of using the correct formulas for heat transfer calculations.
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I have some homework problems that I've been stuck on. Not sure if someone can point me in the right direction...

1) The temperature in an electric oven is 173 °C. The temperature at the outer surface in the kitchen is 39.8 °C. The oven (surface area = 1.51 m^2) is insulated with material that has a thickness of 0.0258 m and a thermal conductivity of 0.03 J/(s m C°). (a) How much energy is used to operate the oven for 1.28 hours? (b) At a price of $0.10 per kilowatt-hour for electrical energy, what is the cost (in dollars) of operating the oven?

I was going to use the equation Q/t=AK delta T/L for the first part but now that I think about it, I think that's the wrong equation to use...

2) The amount of heat per second conducted from the blood capillaries beneath the skin to the surface is 200 J/s. The energy is transferred a distance of 2.4 x 10^-3 m through a body whose surface area is 1.9 m^2. Assuming that the thermal conductivity is that of body fat, determine the temperature difference between the capillaries and the surface of the skin.

I tried using the equation Q=(KA delta T)t/L but once again I think this is wrong

3)A skier wears a jacket filled with goose down that is 15.5 mm thick. Another skier wears a wool sweater that is 7.13 mm thick. Both have the same surface area. Assuming the temperature difference between the inner and outer surfaces of each garment is the same, calculate the ratio of heat lost through wool to heat lost through goose down during the same time interval.

Not even sure what equation to use on this one

4) A car parked in the sun absorbs energy at a rate of 717 watts per square meter of surface area. The car reaches a temperature at which it radiates energy at this same rate. Treating the car as a perfect radiator (e = 1), find the temperature in Kelvin.

I tried to solve this problem using the equation Q/tA=e(5.67x10^-8)(T^4) but I can't get the right answer. Can someone please explain this one step by step. Thanks!
 
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Hi there,

For the first problem, you are correct in using the equation Q = (KAΔT)t/L. To find the energy used to operate the oven for 1.28 hours, we need to first calculate the temperature difference between the inner and outer surfaces of the oven. This can be done by subtracting the outer temperature (39.8 °C) from the inner temperature (173 °C), giving us a ΔT of 133.2 °C. Now, we can plug in the given values into the equation:

Q = (0.03 J/(s m °C)) * (1.51 m^2) * (133.2 °C) * (1.28 hours * 3600 seconds/hour) / (0.0258 m)

This gives us an energy of about 2,104,769.76 J. To find the cost of operating the oven, we can use the formula Cost = (Power * Time * Cost per kWh) / 1000. In this case, the power is given by the energy used (2,104,769.76 J) divided by the time (1.28 hours * 3600 seconds/hour), giving us a power of 490.38 W. Plugging this into the formula, we get a cost of about $0.05.

For the second problem, the equation you are looking for is Q = (KAΔT)/L. Here, we are given the heat conducted per second (200 J/s), the distance (2.4 x 10^-3 m), and the surface area (1.9 m^2). The thermal conductivity of body fat can be found online to be around 0.2 J/(s m °C). Plugging in these values, we get:

200 = (0.2 * 1.9 * ΔT) / (2.4 x 10^-3)

Solving for ΔT, we get a temperature difference of about 20.83 °C.

For the third problem, we can use the equation Q/t = (KAΔT)/L, similar to the first problem. Here, we are trying to find the ratio of heat lost through wool to heat lost through goose down. We can set up two equations, one for each garment, and then divide them to find the ratio. For the wool sweater, we have:

Q1/t
 
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