Concept explainers
(a)
To Calculate:The thermal current in each cube.
(a)
Answer to Problem 37P
The thermal current in copper cube is approximately
The thermal current in aluminum cube is approximately
Explanation of Solution
Given:
Thermal
Areaof cross-section of copper,
Temperature difference,
Thermal conductivity of aluminum,
Areaof cross-section of aluminum,
Formula used: The amount of heat
Here, L the thickness of the substance. The cross-section area of the substance,
temperature difference among the ends of the substance and k the thermal conductivity of thesubstance.
The thermal resistance is given by:
Then the equation (1) is written as
Here, the thermal current
Calculation:
The thermal current within each cube, is expressed as follows:
The thermal resistance of each cube is:
By using the equation (2), the thermal current through the copper can be expressed as follows:
By using the equation (2), the thermal current through the Aluminum can be expressed asfollows:
Hence, the thermal current in aluminum cube is approximately
Conclusion: The thermal current in each cube is to be calculated by using temperature difference and thermal conductivity.
(b)
To Calculate: The total thermal current.
(b)
Answer to Problem 37P
The total thermal current is
Explanation of Solution
Given:Current passing through the copper
Current passing through the aluminum
Formula used:
Given that the cubes are in parallel, therefore total thermal current can be expressed as:
Calculation:
Substitute the values and solve:
Conclusion: Hence,the total thermal current is
(c)
To Calculate:The thermal resistance of two cube combination.
(c)
Answer to Problem 37P
The equivalent thermal resistance is
Explanation of Solution
Given:The temperature difference
Formula used:
The equivalent thermal resistance is obtained by:
Where,
Calculation:
Substitute the values and solve:
Hence, the equivalent thermal resistance is
Conclusion:Equivalent thermal resistance can be calculated using temperature difference and thermal current.
Want to see more full solutions like this?
Chapter 20 Solutions
EBK PHYSICS FOR SCIENTISTS AND ENGINEER
- In 1993, the U.S. government instituted a requirement that all room air conditioners sold in the United States must have an energy efficiency ratio (EER) of 10 or higher. The EER is defined as the ratio of the cooling capacity of the air conditioner, measured in British thermal units per hour, or Btu/h, to its electrical power requirement in watts. (a) Convert the EER of 10.0 to dimensionless form, using the conversion 1 Btu = 1 055 J. (b) What is the appropriate name for this dimensionless quantity? (c) In the 1970s, it was common to find room air conditioners with EERs of 5 or lower. State how the operating costs compare for 10 000-Btu/h air conditioners with EERs of 5.00 and 10.0. Assume each air conditioner operates for 1 500 h during the summer in a city where electricity costs 17.0 per kWh.arrow_forward(a) What is the rate of heat conduction through the 3.00-cm-thick fur of a large animal having a I .40-m surface area? Assume that the animal's skin temperature is 32.0 , that the air temperature is 5.00 , and that has the same thermal conductivity as air. (b) What food intake will the animal need in one day to replace this heat transfer?arrow_forward(a) The inside of a hollow cylinder is maintained at a temperature Ta, and the outside is at a lower temperature, Tb (Fig. P19.45). The wall of the cylinder has a thermal conductivity k. Ignoring end effects, show that the rate of energy conduction from the inner surface to the outer surface in the radial direction is dQdt=2Lk[TaTbln(b/a)] Suggestions: The temperature gradient is dT/dr. A radial energy current passes through a concentric cylinder of area 2rL. (b) The passenger section of a jet airliner is in the shape of a cylindrical tube with a length of 35.0 m and an inner radius of 2.50 m. Its walls are lined with an insulating material 6.00 cm in thickness and having a thermal conductivity of 4.00 105 cal/s cm C. A heater must maintain the interior temperature at 25.0C while the outside temperature is 35.0C. What power must be supplied to the heater? Figure P19.45arrow_forward
- A high-end gas stove usually has at least one burner rated at 14 000 Btu/h. (a) If you place a 0.25-kg aluminum pot containing 2.0 liters of water at 20.C on this burner, how long will it take to bring the water to a boil, assuming all the heat from the burner goes into the pot? (b) Once boiling begins how much time is required to boil all the water out of the pot?arrow_forwardThe surface area of an unclothed person is 1.50 m2, and his skin temperature is 33.0C. The person is located in a dark room with a temperature of 20.0C, and the emissivity of the skin is e = 0.95. (a) At what rate is energy radiated by the body? (b) What is the significance of the sign of your answer?arrow_forwardAn aluminum rod 0.500 m in length and with a cross sectional area of 2.50 cm2 is inserted into a thermally insulated vessel containing liquid helium at 4.20 K. The rod is initially at 3(H) K. (a) If one-halt of the rod is inserted into the helium, how many liters of helium boil off by the time the inserted half cools to 4.20 K? Assume the upper half does not yet cool, (b) If the circular surface of the upper end of the rod is maintained at 300 K. what is the approximate boil-off rate of liquid helium in liters per second after the lower half has reached 4.20 K? (Aluminum has thermal conductivity of 3 100 YV/m K at 4.20 K; ignore its temperature variation. The density of liquid helium is 125 kg/m3.)arrow_forward
- Consider a 6-in * 8-in epoxy glass laminate (k = 0.10 Btu/h·ft·°F) whose thickness is 0.05 in. In order to reduce the thermal resistance across its thickness, cylindrical copper fillings (k = 223 Btu/h·ft·°F) of 0.02 in diameter are to be planted throughout the board, with a center-to-center distance of 0.06 in. Determine the new value of the thermal resistance of the epoxy board for heat conduction across its thickness as a result of this modification.arrow_forwardSuppose a person is covered head to foot by wool clothing with an average thickness of d = 1.95 cm and is transferring energy by conduction through the clothing at the rate of Q / Δt = 45 W. What is the temperature difference, in terms of the quantities given in the problem statement, across the clothing? Denote the surface area of the wool by A and the thermal conductivity by k.arrow_forwardDuring the winter, the inside of an average house is maintained at 20 °C, while the outside temperature is 0 °C. Assuming that the only mechanism of heat transfer is conduction, the walls are 10 cm thick and the heatconductivity of the walls is 0.5 W/(Km). a) Calculate the heat flux from the room to the surroundings in W/m2. b) To reduce the heat loss through the walls, the material should be changed to an insulator material. Thenew overall conductivity will be 0.1 W/(Km); the thickness of the walls is maintained. Calculate the reductionof the heat flux through the walls compared to the initial case.arrow_forward
- The roof of an electrically heated home is 12m long, 16m wide, and 0.50mthick, and is made of a flat layer concrete whose thermal conductivity isk=0.8W/m 2 ·°C. The temperature of the inner and outer surfaces of the roofone night are measured to be 12°C and 1°C, respectively, for a period of 12hours. Determine a) the rate of heat loss through the roof that night b) thecost of that heat loss to the home owner if the cost of electricity is$0.14/kWh.arrow_forwardAn electric kitchen range has a total wall area of 1.40 m2 and is insulated with a layer of fiberglass 4.0 cm thick. The inside surface of the fiberglass has a temperature of 175 degrees C and its outside surface is at 35 degrees C. The fiberglass has a thermal conductivity of 0.040 W/(m*K). What is the heat current through the insulation, assuming it may be treated as a flat slab with an area of 1.40 m2? What electric-power input to the heating element is required to maintain this temperature?arrow_forwardA copper rod has a length of 1.5m1.5m and a cross-sectional area of 4.0×10−4m24.0×10−4m2. One end of the rod is in contact with boiling water and the other with a mixture of ice and water. What is the mass of ice per second that melts? Assume that no heat is lost through the side surface of the rod. Take the thermal conductivity of copper to be k=401W.m.Kk=401W.m.K, and the latent heat of fusion of ice to be 3.35×105J/kg3.35×105J/kg.arrow_forward
- College PhysicsPhysicsISBN:9781285737027Author:Raymond A. Serway, Chris VuillePublisher:Cengage LearningPrinciples of Physics: A Calculus-Based TextPhysicsISBN:9781133104261Author:Raymond A. Serway, John W. JewettPublisher:Cengage LearningPhysics for Scientists and Engineers: Foundations...PhysicsISBN:9781133939146Author:Katz, Debora M.Publisher:Cengage Learning
- Physics for Scientists and Engineers, Technology ...PhysicsISBN:9781305116399Author:Raymond A. Serway, John W. JewettPublisher:Cengage LearningPhysics for Scientists and Engineers with Modern ...PhysicsISBN:9781337553292Author:Raymond A. Serway, John W. JewettPublisher:Cengage Learning