- #1
Randomized10
- 3
- 0
- 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.