Question

Difficulty: MediumModes of Heat Transfer (Conduction, Convection, and Radiation)

A spherical black body radiator with an initial radius of 0.10 m0.10\text{ m} emits thermal radiation at a rate of E1E_1 when its surface temperature is 400 K400\text{ K}. If the radius of the sphere is doubled to 0.20 m0.20\text{ m} and its absolute temperature is reduced to 200 K200\text{ K}, what is the value of the ratio of the new rate of heat radiation to the initial rate, E2E1\frac{E_2}{E_1}?

Answer: 0.25

Answer

The ratio of the new rate of heat radiation to the initial rate is 0.250.25.
According to the Stefan-Boltzmann law, the rate of thermal radiation emitted by a black body is directly proportional to its surface area (Ar2A \propto r^2) and the fourth power of its absolute temperature (T4T^4). Doubling the radius increases the surface area by a factor of 22=42^2 = 4, while halving the absolute temperature decreases the radiation rate per unit area by a factor of (1/2)4=1/16(1/2)^4 = 1/16. The net ratio of the new emission rate to the initial emission rate is 4×(1/16)=0.254 \times (1/16) = 0.25.

Step-by-Step Solution

1
Apply Stefan-Boltzmann's law of radiation to formulate the rate of heat emission.
The total power radiated by a sphere of radius rr at thermodynamic temperature TT is given by P=σAT4=4πσr2T4P = \sigma A T^4 = 4\pi \sigma r^2 T^4, where σ\sigma is the Stefan-Boltzmann constant.
Thermal radiation emission rate depends directly on surface area (A=4πr2A = 4\pi r^2) and the fourth power of absolute temperature (T4T^4).
2
Formulate the ratio E2E1\frac{E_2}{E_1} using the scaled physical parameters.
E2E1=4πσr22T244πσr12T14=(r2r1)2(T2T1)4\frac{E_2}{E_1} = \frac{4\pi \sigma r_2^2 T_2^4}{4\pi \sigma r_1^2 T_1^4} = \left(\frac{r_2}{r_1}\right)^2 \left(\frac{T_2}{T_1}\right)^4
Constants 4πσ4\pi\sigma cancel out when evaluating relative changes.
3
Substitute the given numerical ratios into the equation and compute the result.
Given r2r1=0.200.10=2\frac{r_2}{r_1} = \frac{0.20}{0.10} = 2 and T2T1=200400=0.5\frac{T_2}{T_1} = \frac{200}{400} = 0.5, we get E2E1=(2)2×(0.5)4=4×116=0.25\frac{E_2}{E_1} = (2)^2 \times (0.5)^4 = 4 \times \frac{1}{16} = 0.25.
Evaluating 22=42^2 = 4 and (0.5)4=1/16(0.5)^4 = 1/16 yields a final ratio of 4/16=0.254/16 = 0.25.

Key Concept

Stefan-Boltzmann Law of Radiation
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