Question

Hot water at the temperature of ${77^ \circ }C$ circulates constantly through the tubes. A rom heater is made up of 6 thin – wall tubes of copper, each 1.0 m long and 4.0 cm in diameter. If the emissivity of copper $= 0.8$ and Stefan’s constant $= 5.67 \times {10^{ - 8}}W{m^{ - 2}}{K^{ - 4}}$. Find the amount of heat radiated per second in a room where the average temperature is ${27^ \circ }C$.

Answer 1

Find the radius of tube and then find the total surface area of six tubes. Convert the temperature and radius into their S.I unit. Using the Stefan – Boltzmann Law find the amount of heat radiated per second.

Formula used:
In this question we have to use the Stefan – Boltzmann Law so, the energy emitted per unit time by the body is –
$u = e\sigma A{T^4}$
where, $u$ is the energy emitted by body per unit time
$e$ is the emissivity
$\sigma$ is the Stefan – Boltzmann constant
$A$ is the surface area, and
$T$ is the temperature

Complete step by step solution:
According to the question, it is given that –
Diameter, $d = 4cm$
Length of tube, $l = 1m$
Emissivity of copper, $e = 0.8$
Stefan’s constant, $\sigma = 5.67 \times {10^{ - 8}}W{m^{ - 2}}{K^{ - 4}}$
Let the radius be $r$.
So, the radius is equal to half of the diameter.
$\therefore r = \dfrac{4}{2} = 2cm$
We know that, $1m = 100cm$
$\therefore r = 2 \times {10^{ - 2}}m$
Now, calculating the total surface area of six tubes –
$A = 6\pi rL$
Putting the values of radius and length of tube in the above formula –
$A = 6 \times 2 \times 3.14 \times 2 \times {10^{ - 2}} \times 1 \\\ \Rightarrow A = 0.7536{m^2} \\\$
In the question, it is given –
$T = {77^ \circ }C = 350K$
${T_0} = {27^ \circ }C = 300K$
Now, using the Stefan – Boltzmann’s Law for finding the amount of energy emitted per second. This law describes the power radiated from the body in terms of the temperature.
The amount of heat radiated per second –
$u = e\sigma A\left( {{T^4} - T_0^4} \right)$
Putting the values of emissivity, area of six tubes, temperature and Stefan – Boltzmann constant in the above formula –
$\Rightarrow u = 0.8 \times 5.67 \times {10^{ - 8}} \times 0.7536 \times \left[ {{{\left( {350} \right)}^4} - {{\left( {300} \right)}^4}} \right]$
By doing further calculations, we get –
$\therefore u = 236J/s$

Hence, the amount of heat radiated per second in a room where the average temperature is ${27^ \circ }C$ is $236J/\sec$.

Note: As the shape of tubes is similar to the cylinder so, the area of cylinder can be calculated by the formula –
$\therefore A = 2\pi rh$
Because there are six thin walls tubes of copper, the area of six tubes of copper can be calculated by multiplying six with the area of the cylinder.