Question

. A person is standing (naked) inside their home. If their body temperature is 370 C and the room temperature is 200 C, i. How much power does the person radiate? (4 points) ii. How many calories would the person burn in a day (24 hours) due to this? (4 points)

Answer 1

(i) Power radiated by person is 6793.60 W

(ii) Calories burn by the person in a day is 1.40 x 10⁸

Step-by-step explanation:

Given:

Body temperature of person, T = 370 °C = (370 +273) K = 643 K

Room temperature, T₀ = 200 °C = (200 + 273) K = 473 K

Consider the followings:

Area of the person, A = 2 m²

Emissivity of person, e = 1

Stefan's constant, σ = 5.67 x 10⁻⁸ W/m²K⁴

(i) Power radiated by the person is given by the relation:

P = σeA( T⁴ - T₀⁴)

Substitute the values in the above equation.

`P = 5.67*10^(-8)*1*2(643^(4) -473^(4) )`

P = 6793.60 W

(ii) Consider Q be the energy would burn by the person in a day.

Time, t = 24 hrs = 24 x 60 x 60 s = 86400 s

The relation to determine Q is:

Q = P x t = 6793.60 x 86400

Q = 5.87 x 10⁸ J

But 1 cal = 4.18 J

So, Q = 5.87 x 10⁸/4.18

Q = 1.40 x 10⁸ cal