Question

The robot HooRU is lost in space, floating around aimlessly, and radiates heat into the depths of the cosmos at the rate of 13.1 W. HooRU's surface area is 1.55 m2 and the emissivity of its surface is 0.287. Ignoring the radiation that HooRU absorbs from the cold universe, what is HooRU's temperature T?

Answer 1

Answer:

The temperature is  $T = 168.44 \ K$

Explanation:

From the question ewe are told that

   The rate of heat transferred is    $P = 13.1 \ W$

     The surface area is  $A = 1.55 \ m^2$

      The emissivity of its surface is  $e = 0.287$

Generally, the rate of heat transfer is mathematically represented as

           $H = A e \sigma T^{4}$

=>         $T = \sqrt[4]{\frac{P}{e* \sigma } }$

where  $\sigma$ is the Boltzmann constant with value  $\sigma = 5.67*10^{-8} \ W\cdot m^{-2} \cdot K^{-4}.$

substituting value  

             $T = \sqrt[4]{\frac{13.1}{ 0.287* 5.67 *10^{-8} } }$

            $T = 168.44 \ K$