Question

1. A 170 lbs person burns about 130 kilocalories when running a mile in 8 min. Imagine thirty people (that each weigh 170 lbs) running for one hour at this same pace in a closed gym that is 750ft x 750ft x 20ft. Assuming the air conditioning is broken such that there is no air flow in and out, how much does the air temperature heat up due to the people running for an hour? You can assume that the people stay at constant temperature.

Answer 1

temperature rise of air = 3.14 x 10^-4oC

Explanation

For 1 person

kilocalories burned in 8 min = 130 kcal

kilocalories burned in 1 min = 130 kcal / 8

kilocalories burned in 1 min = 16.25 kcal

kilocalories burned in 60 min = (kilocalories burned in 1 minute) * 60

kilocalories burned in 60 min = (16.25 kcal) * 60

kilocalories burned in 60 min = 975 kcal

For 30 people

kilocalories burned in 60 min = (kilocalories burned in 60 min by 1 person) * 30

kilocalories burned in 60 min = (975 kcal) * 30

kilocalories burned in 60 min = 29250 kcal

kilocalories burned in 60 min (J) = 29250 kcal * (4.184 J / 1 kcal)

kilocalories burned in 60 min (J) = 122382 J

kilocalories burned in 60 min (kJ) = 122.382 kJ

volume of air = 750 ft * 750 ft * 20 ft

volume of air = 11.25 x 10⁶ ft³

volume of air = 11.25 x 10⁶ ft³ * (0.0283168 m³ / 1 ft³)

volume of air = 318564 m³

mass of air = (volume of air) * (density of air)

mass of air = (318564 m³) * (1.225 kg/m³)

mass of air = 390240.9 kg

calories burned by people is absorbed by air which raises temperature of air

heat absorbed by air = (mass of air) * (specific heat of air) * (temperature rise)

temperature rise = heat absorbed by air / [(mass of air) * (specific heat of air)]

temperature rise = 122.382 kJ / [(390240.9 kg) * (1.00 kJ/kg.^oC)]

temperature rise of air = 3.14 x 10^-4oC