1. **State the problem:** We have a coffee temperature function $$f(t) = 56e^{kt} + N$$ where $$f(t)$$ is the temperature at time $$t$$ minutes, and constants $$k$$ and $$N$$ need to be found. 2. **Given information:** The coffee is served at $$78^\circ C$$, so $$f(0) = 78$$. 3. **Use the initial condition:** Substitute $$t=0$$ into the function: $$f(0) = 56e^{k \cdot 0} + N = 56e^0 + N = 56 \times 1 + N = 56 + N$$ 4. **Set equal to initial temperature:** Since $$f(0) = 78$$, $$56 + N = 78$$ 5. **Solve for $$N$$:** $$N = 78 - 56 = 22$$ 6. **Interpretation:** The constant $$N$$ represents the ambient temperature the coffee cools down to, which is $$22^\circ C$$. **Final answer:** $$\boxed{N = 22}$$