JEE Advanced · Physics · 12. Thermal Properties
A human body has a surface area of approximately . The normal body temperature is above the surrounding room temperature . Take the room temperature to be . For , the value of (where is the Stefan-Boltzmann constant). Which of the following options is/are correct?
- A If the surrounding temperature reduces by a small amount , then to maintain the same body temperature the same (living) human being needs to radiate more energy per unit time
- B 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.
- C If the body temperature rises significantly then the peak in the spectrum of electromagnetic radiation emitted by the body would shift to longer wavelengths
- D The amount of energy radiated by the body in 1 second is close to
Answer & Solution
Correct Answer
(C) If the body temperature rises significantly then the peak in the spectrum of electromagnetic radiation emitted by the body would shift to longer wavelengths
Step-by-step Solution
Detailed explanation
\(\text { Energy radiated }=\sigma A\left(T^4-T_0^4\right) \mathrm{t} \)
\( =\sigma A\left[\left(\mathrm{~T}_0+10\right)^4-T_0^4\right] t \)
\( =\sigma A T_0^4\left[\left(1+\frac{10}{T_0}\right)^4-1\right] t \)
\( =\sigma A T_0^4\left[\frac{40}{T_0}\right] \times t=460 \times 1 \times \frac{40}{300} \times 1\) \(=61.33 J \)
\( P=\frac{\text { Energy radiated }}{\text { time }}=\sigma A T^4-\sigma A T_0^4 \)
\( \therefore\left|\frac{d p}{d T_0}\right|=\sigma A\left(4 T_0^3\right) \therefore|d p|=\sigma A\left(4 T_0^3\right) d T_0 \)
\( \therefore|\Delta P|=4 \sigma A T_0^3\)
A, B are not correct options as human body is not a black body. Energy radiated \(\propto \mathrm{A}\) where \(\mathrm{A}\) is the surface area of the body. ' \(\mathrm{C}\) ' is the correct option
\( =\sigma A\left[\left(\mathrm{~T}_0+10\right)^4-T_0^4\right] t \)
\( =\sigma A T_0^4\left[\left(1+\frac{10}{T_0}\right)^4-1\right] t \)
\( =\sigma A T_0^4\left[\frac{40}{T_0}\right] \times t=460 \times 1 \times \frac{40}{300} \times 1\) \(=61.33 J \)
\( P=\frac{\text { Energy radiated }}{\text { time }}=\sigma A T^4-\sigma A T_0^4 \)
\( \therefore\left|\frac{d p}{d T_0}\right|=\sigma A\left(4 T_0^3\right) \therefore|d p|=\sigma A\left(4 T_0^3\right) d T_0 \)
\( \therefore|\Delta P|=4 \sigma A T_0^3\)
A, B are not correct options as human body is not a black body. Energy radiated \(\propto \mathrm{A}\) where \(\mathrm{A}\) is the surface area of the body. ' \(\mathrm{C}\) ' is the correct option
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