The rates of heatradiation from 2 patches of skin each area A , on a patient differ by 2%

If patch of lower temperature is 300 K and emmisivity of both surface is assumed to be unity , the temperature of the other patch will be

Ans 301.5 K

Please give me a detailed simplified solution

To solve the problem of determining the temperature of the second patch of skin, we can use the Stefan-Boltzmann law, which describes how the power radiated by a black body is proportional to the fourth power of its absolute temperature. Since both patches have an emissivity of 1 (which means they behave like perfect black bodies), we can apply this law directly.

Understanding the Problem

We have two patches of skin, both with the same area (A). The lower temperature patch is at 300 K, and the rates of heat radiation from these patches differ by 2%. We need to find the temperature of the second patch, which we will denote as T2.

Applying the Stefan-Boltzmann Law

The Stefan-Boltzmann law states that the power radiated (P) by a black body is given by:

P = εσAT^4

Where:

• P = power radiated
• ε = emissivity (which is 1 for both patches)
• σ = Stefan-Boltzmann constant (approximately 5.67 x 10^-8 W/m²K⁴)
• A = area of the surface
• T = absolute temperature in Kelvin

Setting Up the Equation

Let’s denote the power radiated by the lower temperature patch (300 K) as P1 and the power radiated by the higher temperature patch (T2) as P2. According to the problem, we know:

P2 = P1 + 0.02P1 = 1.02P1

Substituting the expressions for P1 and P2, we have:

σA(300 K)^4 = P1

σA(T2)^4 = P2

Thus, we can write:

σA(T2)^4 = 1.02(σA(300 K)^4)

Simplifying the Equation

Since σ and A are common factors, they can be canceled out from both sides:

(T2)^4 = 1.02(300 K)^4

Calculating T2

Now we need to calculate (300 K)^4:

(300 K)^4 = 300^4 = 8.1 x 10^8 K⁴

Now substituting this value back into the equation:

(T2)^4 = 1.02 * 8.1 x 10^8 K⁴

(T2)^4 = 8.262 x 10^8 K⁴

Finding the Fourth Root

To find T2, we take the fourth root of both sides:

T2 = (8.262 x 10^8 K⁴)^(1/4)

T2 ≈ 301.5 K

Final Result

Thus, the temperature of the other patch of skin is approximately 301.5 K. This calculation illustrates how even a small difference in radiation rates can lead to a measurable difference in temperature, showcasing the sensitivity of thermal radiation principles.