Question

A person eats a dessert that contains 280 Calories. (This "Calorie" unit, with a capital C, is the one used by nutritionists; 1 Calorie = 4186 J.) The skin temperature of this individual is 36°C and that of her environment is 17° C. The emissivity of her skin is 0.75 and its surface area is 1.3 m2. How much time would it take for her to emit a net radiant energy from her body that is equal to the energy contained in this dessert?

Answer 1

Answer:

2.87 hours

Explanation:

The formula of heat transfer by radiation is:

Q = A*e*σ*(T2^4 - T1^4)

where Q is the heat flux, A is the surface area, e is the emissivity, σ = 5.67*10^(-8) W/(m^2*K^4) and T are temperatures. Replacing with data (dimension are omitted):

Q = 1.3*0.75*5.67*10^(-8)*[(273.15 + 36)^4 - (273.15 +17)^4]

Q = 113.156 W or 113.156 J/s

If 1 Calorie is equivalent to 4186 J, then 280 Calories are equivalent to 280*4186 = 1172080 J

If 113.156 J are emitted in 1 second, then 1172080 are emitted in 1172080/113.156 = 10358.0897 seconds or approximately 2.87 hours