Let's solve the problem step by step: First, **identify what is given:** - CO emission rate by Zamboni: 1.1 g/s = 110 mg/s - Volume of rink: 9,000 m³ - Ventilation rate: 1.5 turnovers/hour = 1.5 × 9,000 m³/hour = 13,500 m³/hour - Outdoor CO concentration: 2.5 mg/m³ - Assume steady-state, well-mixed, no reaction. **Step 1: Write the mass balance equation (steady-state):** Rate in (ventilation) + generation = Rate out (ventilation) Let \( C \) = steady-state CO concentration in rink (mg/m³). **Ventilation in:** \( Q \cdot C_{in} \) (mg/hr) \( Q \) = ventilation rate (m³/hr) \( C_{in} \) = 2.5 mg/m³ **Generation:** Zamboni source = 1.1 g/s = 110 mg/s Convert to mg/hr: \( 110 \text{ mg/s} \times 360 \text{ s/hr} = 3,960,000 \text{ mg/hr} \) **Ventilation out:** \( Q \cdot C \) (mg/hr) **Write the equation:** \( (Q \cdot C_{in}) + \text{Generation} = Q \cdot C \) **Plug in values:** \( (13,500 \times 2.5) + 3,960,000 = 13,500 \times C \) **Step 2: Solve for C** Calculate \( 13,500 \times 2.5 = 33,750 \) mg/hr So, \( 33,750 + 3,960,000 = 13,500 \times C \) \( 3,993,750 = 13,500 \times C \) \( C = \frac{3,993,750}{13,500} \) \( C \approx 296 \) mg/m³ **Final Answer:** The steady-state concentration of CO in the air inside the rink will be approximately **296 mg/m³**. This is above the dangerous level of 200 mg/m³, so this situation is hazardous.