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Chapter 14: Problem 67

A black wood stove has a surface area of \(1.20 \mathrm{m}^{2}\) and a surface temperature of \(175^{\circ} \mathrm{C} .\) What is the net rate at which heat is radiated into the room? The room temperature is \(20^{\circ} \mathrm{C}\).

Short Answer

Expert verified
Answer: The net rate at which heat is radiated into the room is approximately 920.83 W.

Step by step solution

01

Convert temperatures to Kelvin

To use the Stefan-Boltzmann Law, we need to convert the given temperatures to Kelvin. The conversion formula is K = C + 273.15. Stove temperature in Kelvin: \(T_s = 175 + 273.15 = 448.15 \mathrm{K}\) Room temperature in Kelvin: \(T_r = 20 + 273.15 = 293.15 \mathrm{K}\)
02

Write the Stefan-Boltzmann Law for power emitted and absorbed

The Stefan-Boltzmann Law states that the power radiated per unit area by a black body is directly proportional to the fourth power of its absolute temperature. \(P = \sigma A T^4\), where \(P\) = radiated power, \(\sigma\) = Stefan-Boltzmann constant \((5.67 \times 10^{-8}\) W m\(^{-2}\) K\(^{-4})\), \(A\) = surface area, and \(T\) = absolute temperature.
03

Calculate the power emitted by the stove

Using the Stefan-Boltzmann Law, we can find the power emitted by the stove. \(P_s = \sigma A_s {T_s}^4\) \(P_s = (5.67 \times 10^{-8}\) W m\(^{-2}\) K\(^{-4})\) × \(1.20 \mathrm{m^2}\) × \((448.15 \mathrm{K})^4\) \(P_s \approx 1241.99 \mathrm{W}\)
04

Calculate the power absorbed by the room

Similarly, we can find the power absorbed by the room. \(P_r = \sigma A_s {T_r}^4\) \(P_r = (5.67 \times 10^{-8}\) W m\(^{-2}\) K\(^{-4})\) × \(1.20 \mathrm{m^2}\) × \((293.15 \mathrm{K})^4\) \(P_r \approx 321.16 \mathrm{W}\)
05

Find the net rate of heat radiation

The net rate of heat radiation is the difference between the power emitted by the stove and the power absorbed by the room. Net rate of heat radiation = \(P_s - P_r\) Net rate of heat radiation = \(1241.99 \mathrm{W} - 321.16 \mathrm{W}\) Net rate of heat radiation \(\approx 920.83 \mathrm{W}\) Thus, the net rate at which heat is radiated into the room is approximately \(920.83 \mathrm{W}\).

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Most popular questions from this chapter

A hotel room is in thermal equilibrium with the rooms on either side and with the hallway on a third side. The room loses heat primarily through a 1.30 -cm- thick glass window that has a height of \(76.2 \mathrm{cm}\) and a width of $156 \mathrm{cm} .\( If the temperature inside the room is \)75^{\circ} \mathrm{F}$ and the temperature outside is \(32^{\circ} \mathrm{F}\), what is the approximate rate (in \(\mathrm{kJ} / \mathrm{s}\) ) at which heat must be added to the room to maintain a constant temperature of \(75^{\circ} \mathrm{F} ?\) Ignore the stagnant air layers on either side of the glass.
(a) What thickness of cork would have the same R-factor as a 1.0 -cm thick stagnant air pocket? (b) What thickness of tin would be required for the same R-factor?
For a cheetah, \(70.0 \%\) of the energy expended during exertion is internal work done on the cheetah's system and is dissipated within his body; for a dog only \(5.00 \%\) of the energy expended is dissipated within the dog's body. Assume that both animals expend the same total amount of energy during exertion, both have the same heat capacity, and the cheetah is 2.00 times as heavy as the dog. (a) How much higher is the temperature change of the cheetah compared to the temperature change of the dog? (b) If they both start out at an initial temperature of \(35.0^{\circ} \mathrm{C},\) and the cheetah has a temperature of \(40.0^{\circ} \mathrm{C}\) after the exertion, what is the final temperature of the dog? Which animal probably has more endurance? Explain.
A stainless steel saucepan, with a base that is made of \(0.350-\mathrm{cm}-\) thick steel \([\kappa=46.0 \mathrm{W} /(\mathrm{m} \cdot \mathrm{K})]\) fused to a \(0.150-\mathrm{cm}\) thickness of copper $[\kappa=401 \mathrm{W} /(\mathrm{m} \cdot \mathrm{K})],\( sits on a ceramic heating element at \)104.00^{\circ} \mathrm{C} .\( The diameter of the pan is \)18.0 \mathrm{cm}$ and it contains boiling water at \(100.00^{\circ} \mathrm{C} .\) (a) If the copper-clad bottom is touching the heat source, what is the temperature at the copper-steel interface? (b) At what rate will the water evaporate from the pan?
If a leaf is to maintain a temperature of \(40^{\circ} \mathrm{C}\) (reasonable for a leaf), it must lose \(250 \mathrm{W} / \mathrm{m}^{2}\) by transpiration (evaporative heat loss). Note that the leaf also loses heat by radiation, but we will neglect this. How much water is lost after 1 h through transpiration only? The area of the leaf is \(0.005 \mathrm{m}^{2}\).
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