For the kangaroo rat a good estimate of thermal conductivity

For the kangaroo rat, a good estimate of thermal conductivity is around 0.01 watts /cm2/K. During the middle of the day, the ground in the Sonoran desert could reach 323 K (50 degrees Celsius). Recall that the k-rat must maintain its body temperature at 311 K and that its surface areas is about 315 cm2.

Conduction= (conductivity) x (Tground-Trat) x (surface area)

Given these values and estimating that about 1/2 the surface of the rat is conducting heat with the ground, calculate how much heat the rat is gaining through conduction.

The problem:

\"If the ground were at 303 K, how much heat would move into the k-rat by conduction? (Answer will be in watts)

Solution

As from Fourier\'s law :

Heat Conduction= (conductivity) x (Tground-Trat) x (surface area)

here conductivity= 0.01 watts /cm²/K.

Tground=323 K

Trat= 311 K

Effective Surface Area=1/2 the surface of the rat ie (1/2) * 315 cm².=157.5 cm²

Putting all the values in above equation we get heat the rat is gaining through conduction

Q=(0.01 watts /cm²/K.)x(323 K-311K)x(157.5 cm²)

=18.9 Watts

If the ground were at 303 K, heat would move into the k-rat by conduction

Q\'=(0.01 watts /cm²/K.)x(311 K-303K)x(157.5 cm²)

=12.6 Watts