Ajogger is running outdoors on a cold day when the temperature is -20.0°c. she is breathing at the rate of 25 breaths per minute, and each time she breathes in she inhales 0.00450 m^3 of air. how much heat does she lose from breathing during 20.0 minutes of jogging if the air in her lungs is heated to her body temperature of 37.0°c before it is exhaled? the specific heat of air is 1020 j/kg ∙ k and the density of air under typical conditions is 1.29 kg/m^3.

168606 J

Explanation:

Amount of heat lost is given by:

Q = m c ΔT

where, m is the mass of air inhaled, c is the specific heat of air and ΔT is the change in the temperature.

First, we will calculate the mass of air inhaled.

In each breath, jogger inhales 0.00450 m³ of air.

Breathing rate = 25 breathes per minute.

She jogs for 20 minutes.

So amount of air inhaled = 25 breathes per minute × 20 minutes × 0.00450 m³

Mass of air inhaled = density of air × volume of air inhaled

⇒ m = 1.29 kg/m³× 25 breathes per minute × 20 minutes × 0.00450 m³  

⇒ m = 2.9 kg

Now, ΔT = T₂-T₁

T₁ = -20⁰C = -20 +273 K = 253 K

T₂ = 37⁰C = 37+273 K = 310 K

⇒ ΔT = T₂-T₁ = 310 K- 253 K = 57 K

Q = m c ΔT

⇒ Q = 2.9 kg × 1020 J/kg.K × 57 K = 168606 J

Thus, the jogger losses 168606 J amount of heat from breathing during 20 minutes of jogging.

0 = 0.40 / 0 = 0

= 0.20 = 9.80 ⁄2 =?

=0 , = = 0.20 =0.50 0 0.40 ⁄

=0−122 =0−12(9.80 ⁄2)(0.50)2 =1.2

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