Question

A metal cube of each side $2\,cm$ long emits $0.16\,kcal$ of heat in $100\,s$. Calculate the emissive power of the cube in joule at that temperature.

Answer 1

Hint: Emissive power is defined as the total radiation emitted or absorbed by a body per unit area and per unit time. Here the total surface area of the cube will be six times the area of each side. We will also convert all units to the units for ease of calculation.
Formula used: Emissive power ${e_p} = \dfrac{Q}{{At}}$.

Complete Step by step answer
The emissive power is given by the formula ${e_p} = \dfrac{Q}{{At}}$,
where ${e_p}$ is the emissive power, $Q$ is the heat absorbed or radiated, $A$ is the surface area of the object in consideration, and $t$ is the time for which the heat radiation has been taking place.
It is given that the heat emitted, $Q$, is equal to $0.16\,kcal$. This when converted to SI units gives $0.16\,k \times 4.2J = 672J$.
The surface area of one side of the cube is
$2cm \times 2cm = 4c{m^2}$.
This when converted to SI units gives us
$4c{m^2} = 4({10^{ - 4}}{m^2}) = 0.0004{m^2}$.
Therefore, the surface area of the entire cube is six times the above value, i.e., $0.0004{m^2} \times 6 = 0.0024{m^2}$
The time during which the heat is emitted is given as $100\,s$.
Thus, substituting the above found values in the formula of emissive power, we get, ${e_p} = \dfrac{Q}{{At}} = \dfrac{{672J}}{{0.0024{m^2} \times 100s}} = 2800\,J/{m^2}s$,
i.e. the emissive power of the metal cube in joules at that temperature is $2800\,J/{m^2}s$.

Therefore, the required answer is $2800\,J/{m^2}s$.

Note: Here the total surface area of the cube will be six times the area of each side as calculated above. We ignore the fact that the cube must be resting on one side and there must be conduction through that side. This problem does not arise in the case of the sphere since the sphere rests at a single point.