- #1

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- Homework Statement
- What power is radiated from a 340°C copper cube 1.0cm on a side? Assume an emissivity of 1

A. 0.76 W

B. 1.8 W

C. 3.4 W

D. 8.0 W

E. 19 W

- Relevant Equations
- $$\frac{dQ}{dt}=e\sigma AT^4$$

##e## is emissivity

##\sigma## is the Stefan-Boltzmann constant, ##5.67*10^{-8} W m^{-2} K^{-4}##

A is the surface area

T is the temperature

##\frac{dQ}{dt}## is the rate of heat transfer or radiated power

At first glance this appeared to be an easy problem, just plug in the values and go, so that's what I did. ##e=1##, ##\sigma=5.67*10^{-8} Wm^{-2}K^{-4}##, ##A=6*(1cm)^2=6*(0.01m)^2=6*0.0001m^2=0.0006m^2=6*10^{-4}m^2##, and ##T=340^{\circ}C=613.15K##. After plugging in the values I got ##\frac{dQ}{dt}=1*5.67*10^{-8}*6*10^{-4}*613.15^4=3.402*10^{-11}*1.4134*10^{11}=4.808W##, but that isn't one of the possible answers no matter how you round it. The answer key says that the correct answer is 19 W, but I don't know how to get there. I tried working out the math using Celsius instead of Kelvin, and got ##\frac{dQ}{dt}=0.4546W##, which isn't right either. Any insights as to what I'm doing wrong would be appreciated.

##\sigma## is the Stefan-Boltzmann constant, ##5.67*10^{-8} W m^{-2} K^{-4}##

A is the surface area

T is the temperature

##\frac{dQ}{dt}## is the rate of heat transfer or radiated power

At first glance this appeared to be an easy problem, just plug in the values and go, so that's what I did. ##e=1##, ##\sigma=5.67*10^{-8} Wm^{-2}K^{-4}##, ##A=6*(1cm)^2=6*(0.01m)^2=6*0.0001m^2=0.0006m^2=6*10^{-4}m^2##, and ##T=340^{\circ}C=613.15K##. After plugging in the values I got ##\frac{dQ}{dt}=1*5.67*10^{-8}*6*10^{-4}*613.15^4=3.402*10^{-11}*1.4134*10^{11}=4.808W##, but that isn't one of the possible answers no matter how you round it. The answer key says that the correct answer is 19 W, but I don't know how to get there. I tried working out the math using Celsius instead of Kelvin, and got ##\frac{dQ}{dt}=0.4546W##, which isn't right either. Any insights as to what I'm doing wrong would be appreciated.