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Chapter 14: Problem 69

At a tea party, a coffeepot and a teapot are placed on the serving table. The coffeepot is a shiny silver-plated pot with emissivity of \(0.12 ;\) the teapot is ceramic and has an emissivity of \(0.65 .\) Both pots hold \(1.00 \mathrm{L}\) of liquid at \(98^{\circ} \mathrm{C}\) when the party begins. If the room temperature is at \(25^{\circ} \mathrm{C},\) what is the rate of radiative heat loss from the two pots? [Hint: To find the surface area, approximate the pots with cubes of similar volume.]

Short Answer

Expert verified
Answer: The rate of radiative heat loss from the two pots at the tea party is 61.186 W.

Step by step solution

01

Calculate the surface area of pots

To find the surface area, we need to approximate the pots with cubes of equivalent volume. The volume of each pot is 1 L or \(1 \times 10^{-3} m^3\). To find the side length of the cube, we use the formula for volume: \(V = a^3\). Therefore, side length of cube: \(a = \sqrt[3]{V} = \sqrt[3]{1 \times 10^{-3}} = 0.1 m\). Now that we know the side length, we can find the surface area of the cube using the formula: \(A = 6a^2\). So the surface area for both cubes (pots) is: \(A = 6(0.1^2) = 0.06 m^2\).
02

Calculate the rate of heat loss for both pots

Now we can calculate the rate of radiative heat loss using the Stefan-Boltzmann equation: \(\frac{dQ}{dt} = e \sigma A (T^4 - T_0^4)\) where e is the emissivity, \(\sigma = 5.67 \times 10^{-8} Wm^{-2}K^{-4}\) is the Stefan-Boltzmann constant, A is the surface area, \(T\) is the initial temperature, and \(T_0\) is the room temperature.
03

Calculate the rate of heat loss for the silver-plated coffeepot

For the silver-plated coffeepot with an emissivity of 0.12, we have: \(\frac{dQ}{dt} = 0.12 \times 5.67 \times 10^{-8} \times 0.06 \times ((98+273)^4 - (25+273)^4) = 11.47 W\)
04

Calculate the rate of heat loss for the ceramic teapot

For the ceramic teapot with an emissivity of 0.65, we have: \(\frac{dQ}{dt} = 0.65 \times 5.67 \times 10^{-8} \times 0.06 \times ((98+273)^4 - (25+273)^4) = 49.716 W\)
05

Add the heat loss rates to find the total heat loss

Now, we add the two heat loss rates calculated above for silver-plated coffeepot and ceramic teapot to find the total heat loss rate: \(\frac{dQ_{total}}{dt} = 11.47 W + 49.716 W = 61.186 W\) The rate of radiative heat loss from the two pots at the tea party is 61.186 W.

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