Answer:
(a) Q = 142.67 W
(b) Basal Metabolic Rate = 178.33 W
Explanation:
(a)
We can find the heat radiated by the person by using Stefan-Boltzman's law:
[tex]Q = \sigma A (T^4 - T_{s}^4)\\[/tex]
where,
Q = heat radiated per second = ?
σ = Stefan-Boltzman Constant = 5.6703 x 10⁻⁸ W/m².k⁴
A = Surface Area = 2 m²
T = Temperature of Skin = 30° C + 273 = 303 k
Ts = Temperature of room = 18° C +273 = 291 k
Therefore,
[tex]Q = (5.6703\ x\ 10^{-8}\ W/m^2.k^4)(2\ m^2)[(303\ k)^4-(291\ k)^4][/tex]
Q = 142.67 W
(b)
Since the heat calculated in part (a) is 80 percent of basal metabolic rate. Therefore,
[tex]Q = (0.8)(Basal\ Metabolic\ Rate)\\Basal\ Metabolic\ Rate = \frac{Q}{0.8}\\\\Basal\ Metabolic\ Rate = \frac{142.67\ W}{0.8}[/tex]
Basal Metabolic Rate = 178.33 W
(a) The excess heat just through radiation Q = 142.67 W
(b) Basal Metabolic Rate = 178.33 W
Basal metabolic rate is the number of calories your body needs to accomplish its most basic (basal) life-sustaining functions.
(a) We can find the heat radiated by the person by using Stefan-Boltzman's law:
[tex]Q=\sigma A(T^4-T_s^4)[/tex]
where,
Q = heat radiated per second = ?
σ = Stefan-Boltzman Constant = 5.6703 x 10⁻⁸ W/m².k⁴
A = Surface Area = 2 m²
T = Temperature of Skin = 30° C + 273 = 303 k
Ts = Temperature of room = 18° C +273 = 291 k
Therefore,
[tex]Q=(5.6703\times 10^{-8}\times (2)\times[(303)-(291)][/tex]
Q = 142.67 W
(b) Since the heat calculated in part (a) is 80 percent of basal metabolic rate. Therefore,
[tex]\rm Q=0.8\times (Basal\ Metabolic\ Rate)[/tex]
[tex]\rm Basal\ metabolic \ Rate = \dfrac{Q}{0.8}[/tex]
[tex]\rm Basal\ Metabolic\ Rate = \dfrac{142.67}{0.8} =178.330W[/tex]
Basal Metabolic Rate = 178.33 W
Hence the excess heat just through radiation Q = 142.67 and Basal Metabolic Rate = 178.33 W
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