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JEE Advanced 2017 · Thermal Properties of Matter

A human body has a surface area of approximately 1 m21\,\text{m}^{2}. The normal body temperature is 10 K10\,\text{K} above the surrounding room temperature T0T_0. Take the room temperature to be T0=300 KT_0 = 300\,\text{K}. For T0=300 KT_0 = 300\,\text{K}, the value of σT04=460 Wm−2\sigma T_0^{4} = 460\,\text{Wm}^{-2} (where σ\sigma is the Stefan-Boltzmann constant). Which of the following options is/are correct?

Answers: 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

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Written solution

Step 1: Identify the given parameters.

The given values for the human body and its surroundings are:

A≈1 m2A \approx 1\ \text{m}^2 T0=300 KT_0 = 300\ \text{K} T=T0+10 K=310 KT = T_0 + 10\ \text{K} = 310\ \text{K} σT04=460 W/m2\sigma T_0^4 = 460\ \text{W/m}^2

Step 2: Evaluate Statement (1) regarding the net rate of heat loss.

The net rate of heat loss by radiation PnetP_{\text{net}} is given by the Stefan-Boltzmann law:

Pnet=σA(T4−T04)P_{\text{net}} = \sigma A (T^4 - T_0^4)

Substitute σ=4603004 W/m2K4\sigma = \frac{460}{300^4}\ \text{W/m}^2\text{K}^4 into the equation:

Pnet=(4603004)(1)(3104−3004)P_{\text{net}} = \left(\frac{460}{300^4}\right) (1) (310^4 - 300^4) Pnet≈5.68×10−8×(9.235×109−8.1×109)P_{\text{net}} \approx 5.68 \times 10^{-8} \times (9.235 \times 10^9 - 8.1 \times 10^9) Pnet≈64.5 W≈60 J/sP_{\text{net}} \approx 64.5\ \text{W} \approx 60\ \text{J/s}

Thus, Statement (1) is correct.

Step 3: Evaluate Statement (2) regarding the change in surrounding temperature.

The net power loss is:

P=σA(T4−T04)P = \sigma A (T^4 - T_0^4)

Differentiating PP with respect to T0T_0 while keeping the body temperature TT constant:

dPdT0=−4σAT03\frac{dP}{dT_0} = -4\sigma A T_0^3

For a small decrease in surrounding temperature ΔT0\Delta T_0, the increase in the rate of heat loss is:

ΔP=4σAT03ΔT0\Delta P = 4\sigma A T_0^3 \Delta T_0

Thus, Statement (2) is correct.

Step 4: Evaluate Statement (3) regarding the reduction of exposed surface area.

The net rate of heat loss is directly proportional to the exposed surface area:

Pnet∝AP_{\text{net}} \propto A

Reducing the exposed surface area (such as by curling up) decreases the rate of heat loss. Thus, Statement (3) is correct.

Step 5: Evaluate Statement (4) regarding the peak wavelength shift.

According to Wien's Displacement Law:

λmax⁡T=constant\lambda_{\max} T = \text{constant} λmax⁡∝1T\lambda_{\max} \propto \frac{1}{T}

As the body temperature TT increases, the peak wavelength λmax⁡\lambda_{\max} shifts to shorter wavelengths. Thus, Statement (4) is incorrect.

Answer: Statements (1), (2), and (3) are correct.

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