Answer
5.0$^\circ$ Celsius
Work Step by Step
Calculate the rate of heat flow using equation 14–5.
$$\frac{Q}{t}=kA\frac{T_1-T_2}{\mathcal{l}}$$
$$95W=(\frac{0.84J}{s \cdot m \cdot C^{\circ}})(4\pi(0.03m)^2)\frac{\Delta T}{0.00050m}$$
Solve for a temperature difference of 5.0 Celsius degrees.