The flux linkage of a closed coil of resistance 2.0 Ω falls from 0.040 Wb to zero…

Step 1. The emfs. $\varepsilon = \dfrac{\Delta(N\Phi)}{\Delta t}$. For $0.20$ s: $\varepsilon = \dfrac{0.040\ \text{Wb}}{0.20\ \text{s}} = 0.20$ V. For $0.80$ s: $\varepsilon = \dfrac{0.040\ \text{Wb}}{0.80\ \text{s}} = 0.050$ V.

Step 2. The charge. $Q = \dfrac{\Delta(N\Phi)}{R} = \dfrac{0.040\ \text{Wb}}{2.0\ \Omega} = 0.020\ \text{C}$, that is $20$ mC. It is the same in both tests.

Step 3. The heat. For $0.20$ s: $H = \dfrac{\varepsilon^2}{R}\,\Delta t = \dfrac{(0.20\ \text{V})^2}{2.0\ \Omega} \times 0.20\ \text{s} = 4.0 \times 10^{-3}\ \text{J}$, that is $4.0$ mJ. For $0.80$ s: $H = \dfrac{(0.050\ \text{V})^2}{2.0\ \Omega} \times 0.80\ \text{s} = 1.0$ mJ.

What we notice. The time is four times longer, so the emf is four times smaller. The charge does not change. The heat is four times smaller, because it goes as $\varepsilon^2\Delta t \propto 1/\Delta t$. The same change of flux costs less energy when it is made slowly.