The top layer of your goose down sleeping bag has a thickness of 5.0 cm and a surface area of 1.0 m^2. When the outside temperature is -18 degrees celcius you lose 24 Cal/hr by heat conduction through the bag (which remains at a cozy 34 degrees celcius inside). Assume that you're sleeping on an insulated pad that eliminates heat conduction to the ground beneath you. What is the thermal conductivity of the goose down?
just for others who are looking for this answer:
You have to convert 24Cal/hr into J/s
since 1Cal=4190 Joules
24Cal/hr x 1hr/3600s x 4190 J = Q/t
Then plug into formula stated above to get a
final answer of 0.026858974 = 0.027 (2 sigfigs)
You must have heard of the basic thermal conduction equation.
dQ/dt = k A (deltaT)/(thickness)
In your case, dQ/dt = 24 Cal/hr
(those are kilocalories, by the way)
deltaT = 52 C
thickness = 0.05 m
A = 1 m^2
Solve for k
is k 0.023076923 ?
the answer is 0.027
I get k = 0.02307.. Cal/hr*degC*m also
Radiation through the sides has been neglected.
To determine the thermal conductivity of the goose down, we can use the formula for heat conduction:
Q = (k * A * (T1 - T2)) / d
Where:
Q is the rate of heat conduction in Cal/hr
k is the thermal conductivity of the material in Cal/(m*hr*C)
A is the surface area of the material in m^2
T1 is the initial temperature in Celsius
T2 is the final temperature in Celsius
d is the thickness of the material in meters
Given values:
Q = 24 Cal/hr
A = 1.0 m^2
T1 = 34 degrees Celsius
T2 = -18 degrees Celsius
d = 5.0 cm = 0.05 m
We want to find the value of k. Rearranging the formula, we can solve for k:
k = (Q * d) / (A * (T1 - T2))
Substituting the given values:
k = (24 Cal/hr * 0.05 m) / (1.0 m^2 * (34 - (-18) degrees Celsius))
Calculating the values:
k = (1.2 Cal) / (1.0 m^2 * 52 degrees Celsius)
k = 0.023 Cal/(m*hr*C)
Therefore, the thermal conductivity of the goose down is approximately 0.023 Cal/(m*hr*C).