Let's solve the problem step by step: **Given:** - Aquarium volume = 1200 L - Initial ammonia concentration = 1 ppm = 1 mg/L = 1.2 g (since 1 ppm = 1 mg/L, 1200 L × 1 mg/L = 1200 mg = 1.2 g) - Dangerous level = 0.25 ppm = 0.25 mg/L = 0.3 g (1200 L × 0.25 mg/L = 300 mg = 0.3 g) - Pump rate = 20 L/min (in and out, so total exchange is 20 L/min) - Fresh water has **zero** ammonia. Let: - \( Q(t) \): amount of ammonia (grams) in the tank at time \( t \) (minutes). --- ### (a) Differential Equation **Outflow rate:** In 1 min, 20 L water leaves, which contains \(\frac{Q(t)}{1200}\) grams/L of ammonia. - Ammonia out per minute: \( 20 \cdot \frac{Q(t)}{1200} = \frac{Q(t)}{60} \) grams/min **Inflow rate:** Fresh water has **0** ammonia (so no ammonia added). **Differential equation:** \[ \frac{dQ}{dt} = \text{input} - \text{output} = 0 - \frac{Q}{60} \] \[ \frac{dQ}{dt} = -\frac{Q}{60} \] **Initial condition:** \( Q(0) = 1.2 \) grams --- ### (b) Time to Reach Safe Level We solve the differential equation: \[ \frac{dQ}{dt} = -\frac{Q}{60} \] This is a separable first-order linear ODE. Integrate: \[ \frac{dQ}{Q} = -\frac{1}{60}dt \] \[ \int \frac{1}{Q} dQ = -\frac{1}{60} \int dt \] \[ \ln |Q| = -\frac{t}{60} + C \] Exponentiate both sides: \[ Q = e^C e^{-t/60} \] Use initial condition \( Q(0) = 1.2 \): \[ 1.2 = e^C \implies C = \ln(1.2) \] So: \[ Q(t) = 1.2 e^{-t/60} \] Set \( Q(t) = 0.3 \) (safe level): \[ 0.3 = 1.2 e^{-t/60} \] \[ \frac{0.3}{1.2} = e^{-t/60} \] \[ 0.25 = e^{-t/60} \] Take natural logarithm: \[ \ln(0.25) = -\frac{t}{60} \] \[ -\ln(4) = -\frac{t}{60} \] \[ \ln(4) = \frac{t}{60} \] \[ t = 60 \times \ln(4) \approx 60 \times 1.386 = 83.2 \text{ minutes} \] --- ### (c) New Pump Rate for 30 Minutes or Less Let the new pump speed be \( r \) L/min. **New ODE:** \[ \frac{dQ}{dt} = -\frac{r}{1200} Q \] So: \[ Q(t) = 1.2 e^{-\frac{r}{1200} t} \] Set \( Q(30) = 0.3 \): \[ 0.3 = 1.2 e^{-\frac{r}{1200} \times 30} \] \[ 0.25 = e^{-\frac{r}{1200} \times 30} \] \[ \ln(0.25) = -\frac{r}{1200} \times 30 \] \[ -1.386 = -\frac{r}{1200} \times 30 \] \[ 1.386 = \frac{r}{1200} \times 30 \] \[ r = \frac{1.386 \times 1200}{30} = \frac{1663.2}{30} \approx 55.44 \text{ L/min} \] --- ## **Final Answers** **(a)** \[ \frac{dQ}{dt} = -\frac{Q}{60}, \quad Q(0) = 1.2 \text{ grams} \] **(b)** It will take about **83.2 minutes** to reach a safe ammonia level (0.25 ppm). **(c)** The new pumps need a flow rate of at least **55.44 L/min** each (if still using two, divide this between two pumps as needed) to reach safe levels in 30 minutes or less.