If a person was locked in a perfectly insulated room 8 ft by 8 ft by 10 ft how long would it take the room temperature to increase from 75 degrees F to 100 degrees F? A person's body heat output is 100 watts and their volume is 3 cubic feet. Also, it is given that the density of air is 1.2 kg/cubic meter and 1 watt sec/(g K) is its specific heat.

Answer 1

Energy to cause temperature change is computed by Q = mC(dT), with Q = change in energy, m = mass, C = specific heat, and dT = temperature change.

Find the volume of the air (remember to exclude the person), and then find the mass of that air, given as 1.2kg/m^3 = 42.384kg/ft^3 (units of volume need to match).

m = (8*8*10 - 3)*(42.384kg/ft^3)

Specific heat is given as 1 W*s/g/K = 1000 W*s/kg/K. (units of mass need to match)

Convert the temperatures to Kelvin to match the specific heat. (Tip: converting to Celsius is sufficient as you are finding a difference between two temperatures.) C = (5/9)(F-32)

T1 = 75F = ?C
T2 = 100F = ?C
dT = T2-T1 = ?

Now we can find Q, the change in energy.
Q = ?

We're given the power output of a person. P = Q/t, so t = Q/P. t = ?

Answer 2

I get that Q, the change in energy is 37498366.5. Using P = Q/t, with P = 100 watt, I get that t = 3749836.665 seconds. The solution is .835 hours. When I divide t by 3600, I don't get the correct answer. Any suggestions?

Answer 3

To answer this type of problem it is very important to include the units at each step.

Volume of box
8 ft x 8 ft x 10 ft = 640 cuft

volume of air = (640-3) cuft = 637 cuft

1 cubic metre = 35.3 cubic feet

volume of air = 637 cuft/35.3 cuft m^-3
= 18.04 m^3

so mass of air = 18.04 m^3 x 1.2 kg m^-3 = 21.65 kg

T1 = 75 degF or 23.9 degC
T2 = 100 degF or 37.8 degC
dT = T2-T1 = 13.9 degC or 13.9 K

specific heat = 1 W s g^-1 K^-1

so energy required to raise temperature by 13.9 K is

energy required is

1 W s g^-1 K^-1
x 21650 g x 13.9 K

= 300935 W s

so time required if power is 100 W is

300935 W s/100 W = 3009.35 s

or 0.836 h

which is close enough to you answer.