Question

Thermal imaging cameras rely on the increase in radiative power to detect temperature variations. A good topical example is the use of camera to detect people suffering with a fever in locations like airports (for example see https://www.youtube.com/watch?v=bESwerHQnVI).

1. [3 marks] For someone of mass 98 kg, how much energy is needed to increase their temperature from T₀ = 37.0 °C to 38.4 °C?
2. [5 marks] The forehead is well known to be a body location that indicates the internal body temperature. For this person, the forehead is at the internal body temperature of 38.3°C, and the tip of nose remains at the room temperature of 25°C. Given that the emissivity of the body is 0.8, how many times more is the radiative intensity (Wm^-2) of the forehead compared to the nose?
3. [5 marks] What is the net radiative intensity (in Wm^-2) of the forehead?
4. [7 marks] The metabolic energy generated in the person is 3.3´10⁶ J in a 6 hour period, and this is balanced by cooling of the body by 0.26 L of perspiration and radiative cooling (you can neglect conductive and convection cooling). What is the effective surface area of the person?

Answer 1

a. The average specific heat of the human body is 3500J/kg so for a 98 kg person the energy required to raise the temp by 1.4C is 1.4*3500*98= 480.2KJ.

b.The nose is at 25C and the forehead at 38.4C. For any given body, the emissive power is . So, Radiative Intensity is Hence, the ratio of Intensity is .

c. Radiative intensity is: . So, .

d. The total energy generated is ` J, 0.26L of perspiration removes 0.26*4186 J= 1088.36 J

    radiative heat transfer is Temperature of the body is 311.3K so, . The value on the left is the heat left after removing the perspiration loss. Therefore effective area = 7744.2/t. The total time is 6*3600= 21600 seconds.

Hence Effective Area =0.3585 m^2.