1. **State the problem:** Calculate the efficiency of a Carnot cycle with a cold reservoir at 0.0 ºC and a hot reservoir at 110 ºC. 2. **Formula:** The efficiency $\eta$ of a Carnot engine is given by $$\eta = 1 - \frac{T_C}{T_H}$$ where $T_C$ and $T_H$ are the absolute temperatures (in Kelvin) of the cold and hot reservoirs respectively. 3. **Convert temperatures to Kelvin:** $$T_C = 0.0 + 273.15 = 273.15\,K$$ $$T_H = 110 + 273.15 = 383.15\,K$$ 4. **Calculate efficiency:** $$\eta = 1 - \frac{273.15}{383.15}$$ 5. **Simplify the fraction:** $$\eta = 1 - \cancel{\frac{273.15}{383.15}} = 1 - 0.713 = 0.287$$ 6. **Round to appropriate significant figures:** Given the temperatures have 3 significant figures, efficiency is $$\eta \approx 0.29$$ **Final answer:** The efficiency of the Carnot cycle is approximately **0.29** (or 29%).