Thermodynamics Problems

Here are the equations for this section:

Try answering the questions and working the problems below to test your understanding of thermodynamics.

1. A can of paint has a warning label that states "Do not store above 40 ^oC". What is this temperature on the Fahrenheit scale?

2. A 1kg block of plastic is heated to 100 ^oC and tossed into 1 kg of water in an insulated cup that is at 20 ^oC. After a while, the two come into equilibrium and the temperture is 40 ^oC. Which has the higher heat capacity, the plastic or water?

3. When you bite into a pizza (extra cheese) straight from the oven, you can seriously burn your mouth. Biting into a peice of bread from the same oven is not so serious. Discuss the possible reasons for the difference.

4. Two sauce pans are on identical stove elements. One pan has 1 kg of water at 0 ^oC while the other has a mixture of ice and water at 0 ^oC with a total mass of 1 kg. Will one take longer than the other to bring to a boil? Why or why not?

5. Consider two identical houses on a hot day. The first house is naturally ventilated and the temperature inside is 85 ^oF. The second house is closed up and an air conditioner keeps the inside temperature at 75 ^oF. Which house will have the greater amount of heat conduction into the house?

6. To reduce heat absorption from the sun, it is wise to paint the exterior of the house a light color with a low value of emissivity. What about the inside of the house? As long as there is little direct sunshine on the walls, does the color make any difference?

   The following problems should be worked sequentially. They put together absorption, heat capacity, radiation and conduction to estimate how a house responds to heating from the sun.

7. The concrete roof of a house is 10 m by 8 m and 10 cm thick (4").
   1. At what rate would the roof absorb energy from the sun if the emissivity of the white roof paint is .08? Does your answer depend upon the time of day?
   2. Estimate the total heat the roof would absorb over the 12 day.

8. Utilizing the answer to the previous problem, estimate the overall or average increase in temperature (DT) of the concrete roof from the energy absorbed from the sun during a 12 hour day. Assume that all of the radiation absorbed goes into heating the roof. The specific heat of concrete is about 900 J/kg, and the density is about 2,300 kg/m³. (The answer is about 8 ^oC.)

   Next we look at radiation and conduction. In the solution to previous problem, we found that the overall temperature of the roof would increase by about 8^oC from the radiation absorption. But the roof will not heat up evenly. The exterior surface of the roof will begin to heat up first and that heat will be conducted to the interior. At the same time, heat will be radiating away from the roof because it will be warmer than the surrounding environment. There are a lot of things happening at once, so we will estimate the extreme values and then discuss the results of all these calculations.

9. At what rate will energy be radiated away from the roof? We need to know both the temperature of the roof and the environment. The values to use are not clear, since the temperatures vary over the day. Let's assume the environment (the air above the roof) is at 27 ^oC (about 80 ^oF) and that there is a breeze that keeps the air just above the roof at this temperature. How hot will the roof be? If all the days heat went into the roof and none of it was radiated or conducted away, then it should never exceed 27 + 8 = 35 ^oC. We have assumed the roof heats up uniformly, so there is not a significant difference between the top and bottom of the roof. (Of course, this can't be completely correct, but we'll see how far off this estimate is when we consider conduction.) Calculate the heat radiated away if the roof is at 35 ^oC.

10. At what rate is heat conducted through the roof to the inside of the house? Using the same logic used in the previous problem, we can argue that the temperature difference between the exterior and interior of the roof shouldn't be more than about 8 ^oC at the warmest part of the day. Find the greatest rate of heat conduction assuming an 8 ^oC difference. (The thermal conductivity of concrete is about .8 W/m^oC.)

11. Consider all the values calculated in the previous sequence of problems and discuss what these results say about the estimations used. What other factors, not already taken into account, might have effected the results? In what direction would those other factors have effected the results?

Review Thermodynamics

Solutions

Don't look at the answers until you've tried the problems on your own!!

1. 40^oC = ⁵/₉ (T_^oF - 32). Solving, we get T_^oF = 40 x ⁹/₅ +32 = 103 ^oF.

2. The water warmed up by 40 - 20 = 20 ^oC while the plastic block cooled down by 100 - 40 = 60 ^oC. Since the mass was the same, water must have a higher heat capacity.

3. There are three possible reasons. The most likely reason is that the specific heat of cheese is much higher than that of the bread. In addition, people have a tendency to put a given volume of food into their mouths. The pizza is probably more dense than the bread and thus has a greater mass than the bread. And finally, although the research on this is scarse, bread probably has a lower thermal conductivity than cheese.

4. It will take longer to boil the ice water mixture. But not because of the temperature difference! That is the same for both cases. The difference is the latent heat of fusion of the ice. It takes 80 cal/gm to melt the ice into water with no change in temperature.

5. The second house! The temperature of the inner wall will be lower and so the temperature difference between outside and inside will be greater.

6. Yes, the interior color does make a difference. The rate at which any surface radiates heat is directly proportional to the emissitivity. As the walls on the inside of the house warm up, they radiate. A darker color radiates more than a lighter color.

   Solutions for the sequenced problems.

7. The maximum absorption occurs at high noon on a clear day. It would be

   ^DQ / _Dt  = (1000 W/m²) (.08) (80) = 6.3 kW

   The actual value would vary from near zero at sunrise to this max value and then back to zero at sunset. The average would be only about half this value.

   Since we want the total heat absorbed for the day, we should use the average absorption, or about 3 kW = 3000 J/s. For 12 hours, that would be a total energy of

   DQ = ^DQ / _Dt x Dt = 3,000 J/s x (12 hrs x 60 min/hr x 60 s/min) = 1.3 x 10⁸ J.

8. Assuming all the heat goes into raising the temperature of the concrete roof, we use DQ  =  m c DT . We need the mass of the roof, m = density x volume = 2,300 x (10 x 8 x .1) = 18,400 kg. So

   1.3 x 10⁸ = (18,400) (900 J/kg^oC) DT

   solving we get DT = 7.9 ^oC. (That's about 14 degrees Fahrenheit.)

9. In the radiation equation, we must use the Kelvin temperture! So T_env = 273 + 27 = 300 ^oK and T_obj = 308 ^oK. Substituting into the equation,
   (^DQ / _Dt )_net  = (.08)(5.67 x 10^-8)(80) (308⁴ - 300⁴)   =   .32 kW

   The maximum amount of heat radiated away, .32 kW, is very small compared to the maximum absorption of 6.3 kW. Even if we double the temperature increase of the roof to 16 ^oC, the heat radiated away is only .679 kW, only about 10% of the absorption rate.

10. We have a temperature difference of DT= 8 ^oC, an area of 80 m² and a thickness of .1 m. Plugging into the heat conduction equation we get

    ^DQ / _Dt  = (.8) (80) ⁽⁸⁾ / _(.1) = 5 kW

    This is nearly the same rate at which heat was being absorbed from the sun's radiation! We will see below why this is not reasonable.

11. Let's list what we have calculated and discuss the assumptions and what factors may have been omitted:
    • Maximum rate of heat absorption from the sun = 6.5 KW and
      total day's heat absorbed from the sun = 1.3 x 10⁸ J.

      The max value is probably reasonably accurate for a clear sunny day ... and the roof is not dirty, which would increase the emissitivity value and the absorption. Using half the max absorption rate to calculate the total day's absorption should be reasonable if it stayed sunny all day.

    • Average temperature increase of roof from absorption = 8 ^oC.

      We did not account for the steel rebar in the concrete. Steel has a much lower heat capacity and would make the temperature increase greater. But rebar usually constitutes only a small percentage of the total mass and our omission is probably not serious.

      The temperature icnrease also assumed that all the heat absorbed stayed in the roof. How good of an assumption was this? That can be answered by looking at how much energy was re-radiated away.

    • Maximum rate of re-radiated heat = .32 kW.

      As mentioned before, this is very small compared to the absorption. We did not include the interior surface, but since the temperature there should be less, it will also radiate less than .32 kW ... unless the interior is painted a dark color. If it were dark, with an emissitivity = .8, then the interior would radiate ten times as much as the outside at the same temperature. This could be significant! It is rare to paint a ceiling dark, but a roof constructed of dark red wood or mahagony is not so rare. Heat radiation is a problem in these cases and plenty of cross ventilation is required to keep the interior of the roof as cool as possible. This type of cooling is convective. The cooler air comes in contact with the roof and carries away heat energy.

    • Maximum rate heat is conducted into the house = 5 kW

      This value is nearly the same as the max absorption value, and is in fact the most erroneous estimate. Recall that we assumed an 8 ^oC temperature difference from outside to inside. At first, this may seem about right, but it can't be! Heat conducted at this rate would mean that essentially all the heat being absorbed was taken into the house. That would leave very little to increase the temperature of the roof. But we got the 8^oC estimate by assuming essentially all the heat stayed in the roof! Consequently, the temperature difference of 8 ^oC was too high. We can estimate the actual temperature difference by realizing that most of the heat leaves the interior of the roof through radiation. That was only about .3 kW at maximum. Since 8 ^oC gave us 5 kW, a simple ratio shows that a temperature difference of about .5 ^oC will give us .3 kW. Note that this means the interior does indeed re-radiate at nearly the same rate as the exterior, assuming similar environmental temperatures. (We would also need to include convective cooling if cooler air was constantly blowing over the interior surface. That would increase the rate of heat conduction to the interior, but reduce the radiation since the surface temperature would be less.)