Heat Transfer in oven

In summary: K)= 0.105 J/m^2K3) For this problem, we can use the same heat transfer equation, but we need to find the thermal conductivity (K) for both goose down and wool. After some research, I found that the thermal conductivity for goose down is 0.026 J/(s m C°) and for wool it is 0.04 J/(s m C°). So, we can plug in these values into the equation:(Q/t)wool = (0.04 J/(s m C°)) x (A) x (delta T) / (L)(Q/t)goose
  • #1
moonlit
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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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  • #2


Dear student,

I am happy to help you with your homework problems. Let's go through each problem one by one and see if we can find the right equations and solutions.

1) For this problem, you are correct in using the equation Q/t=AK delta T/L. This equation is known as the heat transfer equation and it is used to calculate the amount of energy (Q) transferred over a certain time period (t) through a material with a specific thermal conductivity (K), thickness (L), and temperature difference (delta T). In this case, the oven is insulated and we are trying to find the amount of energy used to operate it for 1.28 hours. So, we can plug in the given values into the equation:

Q/t = (0.03 J/(s m C°)) x (1.51 m^2) x (173 °C - 39.8 °C) / (0.0258 m)

= 1584.88 J/s

To find the total amount of energy used in 1.28 hours, we need to convert the time into seconds:

1.28 hours x (60 minutes / 1 hour) x (60 seconds / 1 minute) = 4608 seconds

So, the total energy used in 1.28 hours is:

1584.88 J/s x 4608 seconds = 7,300,022.4 J

(b) To find the cost of operating the oven, we need to convert the energy used into kilowatt-hours (kWh) and then multiply it by the cost per kWh:

7,300,022.4 J x (1 kWh / 3.6x10^6 J) x ($0.10 / 1 kWh) = $2.03

Therefore, it would cost $2.03 to operate the oven for 1.28 hours.

2) For this problem, you are correct in using the equation Q=(KA delta T)t/L. This equation is the same as the heat transfer equation, but rearranged to solve for the temperature difference (delta T). So, we can plug in the given values into the equation:

200 J/s = (K) x (1.9 m^2) x (delta T) / (2.4x10^-3 m)

Solving for delta T, we get:

delta T = (200 J/s) x (2.
 
  • #3


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
 

1. How does heat transfer occur in an oven?

In an oven, heat transfer occurs through convection, conduction, and radiation. Convection is the transfer of heat through the movement of hot air or fluids, such as when hot air circulates inside an oven. Conduction is the transfer of heat through direct contact, such as when the food touches the hot oven walls or racks. Radiation is the transfer of heat through electromagnetic waves, such as when the heating element emits infrared radiation to heat up the food.

2. What is the role of the heating element in an oven?

The heating element in an oven is responsible for producing heat through electrical resistance. When the heating element is turned on, it converts electrical energy into heat energy, which is then transferred to the food through radiation.

3. How does the type of material of the baking dish affect heat transfer in an oven?

The type of material of the baking dish can affect heat transfer in an oven. Materials that are good conductors of heat, such as metal, can transfer heat more efficiently and evenly compared to materials that are poor conductors of heat, such as glass. This can affect the cooking time and the quality of the food.

4. What is the best oven setting for even heat distribution?

The best oven setting for even heat distribution depends on the type of food you are cooking. For baking, setting the oven to "bake" mode is recommended as it uses both the top and bottom heating elements, providing even heat distribution. For roasting, setting the oven to "convection roast" mode, which uses a fan to circulate hot air, can also result in even heat distribution.

5. How can I prevent uneven heat distribution in my oven?

To prevent uneven heat distribution in your oven, make sure to preheat the oven to the desired temperature before placing the food inside. Avoid overcrowding the oven with too many dishes, as this can block the flow of hot air and result in uneven heating. You can also rotate the baking dish halfway through the cooking time to ensure even heat distribution.

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