Chapter 1: Problem 46
Define emissivity and absorptivity. What is Kirchhoff's law of radiation?
Chapter 1: Problem 46
Define emissivity and absorptivity. What is Kirchhoff's law of radiation?
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Get started for freeA cold bottled drink ( $\left.m=2.5 \mathrm{~kg}, c_{p}=4200 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\( at \)5^{\circ} \mathrm{C}$ is left on a table in a room. The average temperature of the drink is observed to rise to \(15^{\circ} \mathrm{C}\) in \(30 \mathrm{~min}\). The average rate of heat transfer to the drink is (a) \(23 \mathrm{~W}\) (b) \(29 \mathrm{~W}\) (c) \(58 \mathrm{~W}\) (d) \(88 \mathrm{~W}\) (e) \(122 \mathrm{~W}\)
An ice skating rink is located in a building where the air is at $T_{\text {air }}=20^{\circ} \mathrm{C}\( and the walls are at \)T_{w}=25^{\circ} \mathrm{C}$. The convection heat transfer coefficient between the ice and the surrounding air is \(h=10 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The emissivity of ice is \(\varepsilon=0.95\). The latent heat of fusion of ice is \(h_{i f}=333.7 \mathrm{~kJ} / \mathrm{kg}\), and its density is $920 \mathrm{~kg} / \mathrm{m}^{3}$. (a) Calculate the refrigeration load of the system necessary to maintain the ice at \(T_{s}=0^{\circ} \mathrm{C}\) for an ice rink of \(12 \mathrm{~m}\) by \(40 \mathrm{~m}\). (b) How long would it take to melt \(\delta=3 \mathrm{~mm}\) of ice from the surface of the rink if no cooling is supplied and the surface is considered insulated on the back side?
Can all three modes of heat transfer occur simultaneously (in parallel) in a medium?
A \(0.3\)-cm-thick, \(12-\mathrm{cm}\)-high, and \(18-\mathrm{cm}\)-long circuit board houses 80 closely spaced logic chips on one side, each dissipating $0.06 \mathrm{~W}$. The board is impregnated with copper fillings and has an effective thermal conductivity of $16 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}$. All the heat generated in the chips is conducted across the circuit board and is dissipated from the back side of the board to the ambient air. Determine the temperature difference between the two sides of the circuit board. Answer: \(0.042^{\circ} \mathrm{C}\)
It is well known that at the same outdoor air temperature a person is cooled at a faster rate under windy conditions than under calm conditions due to the higher convection heat transfer coefficients associated with windy air. The phrase wind-chill is used to relate the rate of heat loss from people under windy conditions to an equivalent air temperature for calm conditions (considered to be a wind or walking speed of \(3 \mathrm{mph}\) or $5 \mathrm{~km} / \mathrm{h}$ ). The hypothetical wind-chill temperature (WCT), called the wind-chill temperature index (WCTI), is an air temperature equivalent to the air temperature needed to produce the same cooling effect under calm conditions. A 2003 report on wind-chill temperature by the U.S. National Weather Service gives the WCTI in metric units as WCTI $\left({ }^{\circ} \mathrm{C}\right)=13.12+0.6215 T-11.37 \mathrm{~V}^{0.16}+0.3965 T V^{0.16}$ where \(T\) is the air temperature in \({ }^{\circ} \mathrm{C}\) and \(V\) the wind speed in \(\mathrm{km} / \mathrm{h}\) at \(10 \mathrm{~m}\) elevation. Show that this relation can be expressed in English units as WCTI $\left({ }^{\circ} \mathrm{F}\right)=35.74+0.6215 T-35.75 V^{0.16}+0.4275 T V^{0.16}$ where \(T\) is the air temperature in \({ }^{\circ} \mathrm{F}\) and \(V\) the wind speed in \(\mathrm{mph}\) at \(33 \mathrm{ft}\) elevation. Also, prepare a table for WCTI for air temperatures ranging from 10 to \(-60^{\circ} \mathrm{C}\) and wind speeds ranging from 10 to \(80 \mathrm{~km} / \mathrm{h}\). Comment on the magnitude of the cooling effect of the wind and the danger of frostbite
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