Many modern technologies, such as computers and smartphones, are built on the principles of Physics Topics such as quantum mechanics and information theory.
How Will You Derive Newton’s Law of Cooling From Stefan’s Law?
Statement: The rate of loss of heat by a body, due to radiation, is directly proportional to the temperature difference between the body and its surroundings, provided the temperature difference is small.
Time of Cooling: Let a body, of mass m and specific heat s, at a temperature θ be placed in a surrounding of temperature θ0, where θ0 < θ.
According to Newton’s law of cooling, the rate of loss of heat by the body, \(\frac{d Q}{d t}\) ∝ -(θ – θ0)
Also from calorimetry, \(\frac{d Q}{d t}\) = ms \(\frac{d \theta}{d t}\) where \(\frac{d \theta}{d t}\) = rate of fall of temperature.
∴ \(\frac{d \theta}{d t}\) ∝ -(θ – θ0) or, \(\frac{d \theta}{d t}\) = -C(θ – θ0)
where C is a constant and the -ve sign indicates fall in temperature with time.
∴ \(\frac{d \theta}{\theta-\theta_0}\) = -C dt
By integration, \(\int_{\theta_1}^{\theta_2} \frac{d \theta}{\theta-\theta_0}\) = – C\(\int_0^t\) dt,
[At the time of cooling the temperature of the body decreases from θ1 to θ2 in time t, where θ1 > θ2]
Newton’s law of cooling from Stefan-Boltzmann law: When the difference in temperature between the body and its surroundings is low, Stefan-Boltzmann law becomes equivalent to Newton’s law of cooling. Let the absolute temperature of the body and that of the surroundings be T and T0 respectively. (T – T0) is very small.
From Stefan’s law, heat radiation per unit area per unit time
E = σ(T4 – \(T_0^4\)) = σ(T – T0)(T3 + T2T0 + T\(T_0^2\) + \(T_0^3\))
If (T – T0) is very small, i.e., if T \(\simeq\) T0, we can write the equation as,
E = σ(T – T0) × 4\(T_0^3\) = C(T – T0), where C = 4σ\(T_0^3\)
At a constant surrounding temperature, C is a constant
Hence, E ∝ (T – T0), which is Newton’s law of cooling.
The body will cool by radiation following Newton’s law of cooling only when temperature difference between the body
and its surroundings is small.
Newton’s law of cooling is valid for cooling by conduction [see equation (1) in section 9.2.1], when k, A and d are constants. It is also valid for some cases of cooling by convection.
Numerical Examples
Example 1.
The initial temperature of a body is 353 K. It reduces to 337 K in 5 mins and to 325 K in 10 mins. What will be its temperature after 15 mins? What is the temperature of the surroundings?
Solution:
From Newton’s law of cooling we have,
ln\(\frac{\theta_1-\theta_0}{\theta_2-\theta_0}\) = Ct
After 5 mins, ln\(\frac{353-\theta_0}{337-\theta_0}\) = 5C ……………. (1)
After the next 5 mins, ln\(\frac{337-\theta_0}{325-\theta_0}\) = 5C ………… (2)
Yet After another 5 mins, ln\(\frac{325-\theta_0}{\theta_3-\theta_0}\) = 5C …… (3)
[where θ3 is the temperature of the body after 15 mins]
From equation (1) and (2) we have,
\(\frac{353-\theta_0}{337-\theta_0}\) = \(\frac{337-\theta_0}{\theta_3-\theta_0}\) or, θ0 = 289 K
∴ Temperature of the surroundings = 289 K.
From equations (2) and (3) we have,
\(\frac{337-\theta_0}{325-\theta_0}\) = \(\frac{325-\theta_0}{\theta_3-\theta_0}\)
or, \(\frac{337-289}{325-289}\) = \(\frac{325-289}{\theta_3-289}\) or, θ3 = 316 K
Example 2.
The temperature of a black body is increased from 27°C to 927°C. What is the ratio of the heat radiated at these two temperatures?
Solution:
T1 = 27°C = (27 + 273) K = 300 K ;
T2 = 927°C = (927 + 273) K = 1200 K
According to Stefan’s law,
E ∝ T4
or, \(\frac{E_1}{E_2}\) = \(\left(\frac{T_1}{T_2}\right)^4\) = \(\left(\frac{300}{1200}\right)^4\) = \(\left(\frac{1}{4}\right)^4\) = \(\frac{1}{256}\).
Example 3.
A spherical black body of radius 12 cm radiates 450 W heat energy at 500 K. Now, if the radius is reduced to half and the temperature doubled, what will be the amount of heat radiated?
Solution:
The surface area of a sphere, A = πr2, i.e., A ∝ r2
Rate of radiation from the surface of a black body,
E ∝ AT4 or, E ∝ r2T4 [∵ A ∝ r2]
∴ \(\frac{E_1}{E_2}\) = \(\left(\frac{r_1}{r_2}\right)^2\left(\frac{T_1}{T_2}\right)^4\)
or, E2 = E1 × \(\left(\frac{r_2}{r_1}\right)^2\left(\frac{T_2}{T_1}\right)^4\) = 450 × \(\left(\frac{1}{2}\right)^2\) × \(\left(\frac{2}{1}\right)^4\)
= 1800 W.
Example 4.
The filament of an electric bulb of 40 W has an average temperature of 2500°C. If the length and diameter of the filament are 10 cm and 0.1 mm respectively, then determine its coefficient of radiation (ϵ), provided all the heat is emitted only through radiation, Given, σ = 5.67 × 10-5 CGS unit.
Solution:
We know, \(\frac{Q}{t}\) = ϵAσT4
Here, \(\frac{Q}{t}\) = radiation per second = 40 W = 40 × 107 erg ᐧ s-1
A = π ld = π × 10 × 0.01 = 0.1π cm2
σ = 5.67 × 10-5 CGS unit, T = 2500 + 273 = 2773 K.
∴ 40 × 107 = ϵ × 0.1π × 5.67 × 10-5 × (2773)4
or, ϵ = 0.38 cal ᐧ °C-2 ᐧ cm-2
Example 5.
With a body temperature of 37°C, a person of body surface area 1.40 m2 and coefficient of emission 0.85, stands in a room kept at a temperature of 20°C. If Stefan’s constant, σ = 5.67 × 10-8 W ᐧ m-2 ᐧ K-4, find the rate of loss of heat by the man by radiation only.
Solution:
Heat lost by the man per min is given by,
Q = ϵAσ(T4 – \(T_0^4\)) × t
= 0.85 × 1.40 × 5.67 × 10-8[3104 – 2934] × 60
= 7550 J = 1800 cal (approx.)
Example 6.
The earth receives radiation from the sun at the rate of 1400 W ᐧ m-2. The distance of the centre of the sun from the surface of the earth is 1.5 × 1011m and the radius of the sun is 7 × 108m. Treating the sun as a black body, determine its surface temperature in Kelvin scale. σ = 5.67 × 10-8 W ᐧ m-2 ᐧ K-4
Solution:
If T be the surface temperature of the sun, then
T = \(\left[\left(\frac{R}{r}\right)^2 \times \frac{S}{\sigma}\right]^{1 / 4}\)
Here, S = 1400 W ᐧ m-2, R = 1.5 × 1011m, r = 7 × 108m, σ = 5.67 × 10-8 W ᐧ m-2 ᐧ K-4
∴ T = \(\left[\frac{\left(1.5 \times 10^{11}\right)^2}{\left(7 \times 10^8\right)^2} \times \frac{1400}{5.67 \times 10^{-8}}\right]^{1 / 4}\)
= [0.1128 × 1016]1/4 = 5800K