A human body has a surface area of approximately $1 \mathrm{~m}^{2}$. The normal body temperature is 10 K above the surrounding room temperature $T_{0}$. Take the room temperature to be $T_{0}=300 \mathrm{~K}$. For $T_{0}=300 \mathrm{~K}$, the value of $\sigma T_{0}^{4}=460 \mathrm{Wm}^{-2}$ (where $\sigma$ is the Stefan-Boltzmann constant). Which of the following options is/are correct?

(C) Reducing the exposed surface area of the body (e.g. by curling up) allows humans to maintain the same body temperature while reducing the energy lost by radiation

(A) The amount of energy radiated by the body in 1 second is close to 60 Joules

(B) If the surrounding temperature reduces by a small amount $\Delta T_{0} \ll T_{0}$, then to maintain the same body temperature the same (living) human being needs to radiate $\Delta W=4 \sigma T_{0}^{3} \Delta T_{0}$ more energy per unit time

(D) If the body temperature rises significantly then the peak in the spectrum of electromagnetic radiation emitted by the body would shift to longer wavelengths

Heat & Thermodynamics — JEE Advanced · 2017 · 21 May 2017 · Paper 1 Question with Solution

A human body has a surface area of approximately

1 m^{2}

. The normal body temperature is 10 K above the surrounding room temperature

T_{0}

. Take the room temperature to be

T_{0} = 300 K

. For

T_{0} = 300 K

, the value of

σ T_{0}^{4} = 460 Wm^{- 2}

(where

σ

is the Stefan-Boltzmann constant). Which of the following options is/are correct?

(A)The amount of energy radiated by the body in 1 second is close to 60 Joules
(B)If the surrounding temperature reduces by a small amount $Δ T_{0} ≪ T_{0}$ , then to maintain the same body temperature the same (living) human being needs to radiate $Δ W = 4 σ T_{0}^{3} Δ T_{0}$ more energy per unit time
(C)Reducing the exposed surface area of the body (e.g. by curling up) allows humans to maintain the same body temperature while reducing the energy lost by radiation✓ Correct
(D)If the body temperature rises significantly then the peak in the spectrum of electromagnetic radiation emitted by the body would shift to longer wavelengths

Answer: (C)

Solution

The net power radiated by the body is

P = σ A (T_{b o d y}^{4} - T_{0}^{4}) = σ A [(T_{0} + 10)^{4} - T_{0}^{4}]

. Using the binomial approximation for small

Δ T

(T_{0} + 10)^{4} \approx T_{0}^{4} + 4 T_{0}^{3} \cdot 10

. Thus

P \approx σ A \cdot 4 T_{0}^{3} \cdot 10 = 4 \times 460 \times 1 \times 10 = 18400

W, not 60 J/s, so option (A) is incorrect. For option (B), if

T_{0}

decreases, the body still maintains the same temperature, so the radiated power increases, but the formula given applies to the change in absorbed power, not radiated power, making (B) incorrect. Option (C) is correct: reducing surface area

A

reduces the total power radiated

P = σ A (T_{b o d y}^{4} - T_{0}^{4})

. By Wien's law,

λ_{m} T = constant

, so if body temperature rises, the peak wavelength shifts to shorter wavelengths, making option (D) incorrect.

JEE Advanced · 2017 · 21 May 2017 · Paper 1 · One or more correct options

JEE Advanced · 2017 · 21 May 2017 · Paper 1 · Heat & Thermodynamics

Solution

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