A magnet of mass 0.082 kg falls through a copper tube 0.90 m long, and reaches a steady speed of…

Step 1. The drag constant. At a steady speed $kv_t = mg$. We want $k$, so $k = \dfrac{mg}{v_t} = \dfrac{0.082\ \text{kg} \times 9.8\ \text{m/s}^2}{0.085\ \text{m/s}} = 9.45\ \text{kg/s}$.

Step 2. The heat each second. $P = mgv_t = 0.804\ \text{N} \times 0.085\ \text{m/s} = 0.0683\ \text{W}$, that is $68.3$ mW.

Step 3. The time. $t = \dfrac{h}{v_t} = \dfrac{0.90\ \text{m}}{0.085\ \text{m/s}} = 10.6\ \text{s}$.

Step 4. The total heat. The magnet ends with the same speed that it began with. All the potential energy it loses becomes heat: $mgh = 0.082\ \text{kg} \times 9.8\ \text{m/s}^2 \times 0.90\ \text{m} = 0.723\ \text{J}$.