Question

The thermal conductivity of the fur (including the skin) of a male Husky dog is \(0.026 \mathrm{W} /(\mathrm{m} \cdot \mathrm{K}) .\) The dog's heat output is measured to be \(51 \mathrm{W}\), its internal temperature is $38^{\circ} \mathrm{C},\( its surface area is \)1.31 \mathrm{m}^{2},$ and the thickness of the fur is \(5.0 \mathrm{cm} .\) How cold can the outside temperature be before the dog must increase its heat output?

Short answer

The minimum outside temperature before the dog must increase its heat output is approximately -36.9°C.

Step-by-step solution

Convert the thickness of the fur to meters

The thickness of the fur is given in centimeters. Convert it to meters: \(d = 5.0 \mathrm{cm} = 0.05 \mathrm{m}\)

Identify the known variables

The known variables are: - Thermal conductivity, \(k = 0.026 \mathrm{W} /(\mathrm{m} \cdot \mathrm{K})\) - Heat output, \(Q = 51 \mathrm{W}\) - Internal temperature, \(T_{hot} = 38^{\circ}\mathrm{C}\) - Surface area, \(A = 1.31 \mathrm{m}^2\) - Thickness of the fur, \(d = 0.05 \mathrm{m}\)

Rearrange the heat transfer formula to solve for \(T_{cold}\)

We will rearrange the formula to solve for the temperature of the colder side: \(T_{cold} = T_{hot} - \frac{Q * d}{k * A}\)

Substitute the known values and solve for \(T_{cold}\)

Now, plug the known values into the formula and solve for the outside temperature: \(T_{cold} = 38 - \frac{51 * 0.05}{0.026 * 1.31} = 38 - \frac{2.55}{0.03406} \approx 38 - 74.9\) \(T_{cold} \approx -36.9^{\circ}\mathrm{C}\) So, the outside temperature can be as low as approximately \(-36.9^{\circ}\mathrm{C}\) before the Husky dog must increase its heat output.