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An exterior wall is \(5.869 \mathrm{~m}\) wide and \(3.289 \mathrm{~m}\) tall, and \(69.71 \mathrm{~W}\) of power is carried through it. The outdoor temperature is \(3.857{ }^{\circ} \mathrm{C},\) and the indoor temperature is \(24.21^{\circ} \mathrm{C} .\) What is the \(R\) factor of the material with which the wall is insulated?

Short Answer

Expert verified
Answer: The R factor of the material is 5.5934.

Step by step solution

01

Calculate the Area of the wall

To calculate the area, we will multiply the width and the height. The formula for the area (A) of a rectangle is: A = width * height In this case, width = 5.869 m, and height = 3.289 m. Therefore, the area is: A = 5.869 * 3.289 = 19.306641 \mathrm{~m^2}
02

Calculate the Temperature Difference

Next, we need to find the temperature difference between the indoor and outdoor environments. To do this, we will subtract the outdoor temperature from the indoor temperature. The formula for the temperature difference (ΔT) is: ΔT = T_inside - T_outside The indoor temperature is 24.21 °C, and the outdoor temperature is 3.857 °C: ΔT = 24.21 - 3.857 = 20.353 { }^{\circ} \mathrm{C}
03

Calculate the R Factor

Now that we have the area and the temperature difference, we can find the R factor. We will use the formula mentioned earlier: R = (Area * Temperature Difference) / Power We have the area (A = 19.306641 m²), the temperature difference (ΔT = 20.353 °C), and the power (P = 69.71 W): R = (19.306641 * 20.353) / 69.71 = 5.5934 Hence, the R factor of the material with which the wall is insulated is 5.5934.

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Most popular questions from this chapter

A 1.19-kg aluminum pot contains \(2.31 \mathrm{~L}\) of water. Both pot and water are initially at \(19.7^{\circ} \mathrm{C} .\) How much heat must flow into the pot and the water to bring their temperature up to \(95.0^{\circ} \mathrm{C}\) ? Assume that the effect of water evaporation during the heating process can be neglected and that the temperature remains uniform throughout the pot and the water.

In the isothermal compression of a gas, the volume occupied by the gas is decreasing, but the temperature of the gas remains constant. In order for this to happen, a) heat must enter the gas. b) heat must exit the gas. c) no heat exchange should take place between the gas and the surroundings.

An exterior wall is \(5.183 \mathrm{~m}\) wide and \(3.269 \mathrm{~m}\) tall. The wal is insulated with a material that has an \(R\) factor of \(29 .\) The outdoor temperature is \(1.073^{\circ} \mathrm{C},\) and the indoor temperature is \(23.37^{\circ} \mathrm{C} .\) How much power is carried through the wall?

Which of the following does not radiate heat? a) ice cube b) liquid nitrogen c) liquid helium d) a device at \(T=0.010 \mathrm{~K}\) e) all of the above f) none of the above

The radiation emitted by a blackbody at temperature \(T\) has a frequency distribution given by the Planck spectrum: $$\epsilon_{T}(f)=\frac{2 \pi h}{c^{2}}\left(\frac{f^{3}}{e^{h f / k_{\mathrm{B}} T}-1}\right)$$ where \(\epsilon_{T}(f)\) is the energy density of the radiation per unit increment of frequency, \(v\) (for example, in watts per square meter per hertz), \(h=6.626 \cdot 10^{-34} \mathrm{~J} \mathrm{~s}\) is Planck's constant, \(k_{\mathrm{B}}=1.38 \cdot 10^{-23} \mathrm{JK}^{-1}\) is the Boltzmann constant, and \(c\) is the speed of light in vacuum. (We'll derive this distribution in Chapter 36 as a consequence of the quantum hypothesis of light, but here it can reveal something about radiation. Remarkably, the most accurately and precisely measured example of this energy distribution in nature is the cosmic microwave background radiation.) This distribution goes to zero in the limits \(f \rightarrow 0\) and \(f \rightarrow \infty\) with a single peak between those limits. As the temperature is increased, the energy density at each frequency value increases, and the peak shifts to a higher frequency value. a) Find the frequency corresponding to the peak of the Planck spectrum, as a function of temperature. b) Evaluate the peak frequency at \(T=6.00 \cdot 10^{3} \mathrm{~K}\), approximately the temperature of the photosphere (surface) of the Sun. c) Evaluate the peak frequency at \(T=2.735 \mathrm{~K}\), the temperature of the cosmic background microwave radiation. d) Evaluate the peak frequency at \(T=300 . \mathrm{K},\) which is approximately the surface temperature of Earth.

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