Power Radiated From a Copper Cube

[Randomized10]

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.

 

[kuruman]

My calculation agrees with yours. This is not the first time that the correct solution does not match any of the offered answers. I suggest that you show your answer to the person who assigned you this problem because he/she/ze/they needs to know.

 

[TSny]

I agree with @kuruman. Your answer of 4.8 W looks correct. The answer of 19 W corresponds to an edge length of 2.0 cm.

 

[Randomized10]

Ok, I'll talk to the professor. Thank you!