Solution
(a) Time to reach boiling point
Phase 1: Heating without inflow
Given: $V_i = 20\,\text{L}$, $\theta_i = 20^\circ\text{C}$. The water is heated to $\theta_1 = 60^\circ\text{C}$ in time $\Delta t_1 = 40\,\text{min}$.
Since heat loss and vessel capacity are negligible, the power of the heater $P$ is:
$$ P = \frac{\text{Heat supplied}}{\text{Time}} = \frac{m s \Delta \theta}{\Delta t_1} = \frac{V_i \rho s (\theta_1 – \theta_i)}{\Delta t_1} $$
Phase 2: Heating with constant inflow
Now, water at $\theta_i = 20^\circ\text{C}$ is added at a rate $r = 200\,\text{cm}^3/\text{min}$. Let $t$ be the additional time required to reach the boiling temperature $\theta_b = 100^\circ\text{C}$.
During time $t$, total energy supplied by the heater must equal the energy required to raise the initial water volume to $\theta_b$ plus the energy required to raise the added water to $\theta_b$.
Energy Balance Equation:
$$ P \cdot t = \underbrace{V_i \rho s (\theta_b – \theta_1)}_{\text{Initial water heating}} + \underbrace{(r \cdot t) \rho s (\theta_b – \theta_i)}_{\text{Added water heating}} $$
Substituting the expression for $P$:
$$ \left[ \frac{V_i \rho s (\theta_1 – \theta_i)}{\Delta t_1} \right] t = V_i \rho s (\theta_b – \theta_1) + r t \rho s (\theta_b – \theta_i) $$
Canceling $\rho s$ from both sides:
$$ \frac{V_i (\theta_1 – \theta_i)}{\Delta t_1} t – r t (\theta_b – \theta_i) = V_i (\theta_b – \theta_1) $$
$$ t \left[ \frac{V_i (\theta_1 – \theta_i)}{\Delta t_1} – r (\theta_b – \theta_i) \right] = V_i (\theta_b – \theta_1) $$
Solving for $t$:
$$ t = \frac{V_i (\theta_b – \theta_1)}{\frac{V_i (\theta_1 – \theta_i)}{\Delta t_1} – r (\theta_b – \theta_i)} = \frac{V_i (\theta_b – \theta_1) \Delta t_1}{V_i (\theta_1 – \theta_i) – r \Delta t_1 (\theta_b – \theta_i)} $$
The total time from the beginning is $T_{total} = \Delta t_1 + t$:
$$ T_{total} = \Delta t_1 \left[ 1 + \frac{V_i (\theta_b – \theta_1)}{V_i (\theta_1 – \theta_i) – r \Delta t_1 (\theta_b – \theta_i)} \right] $$
Numerical Substitution:
- $V_i = 20\,\text{L}$, $r = 0.2\,\text{L/min}$
- $\theta_i = 20^\circ\text{C}$, $\theta_1 = 60^\circ\text{C}$, $\theta_b = 100^\circ\text{C}$
- $\Delta t_1 = 40\,\text{min}$
Calculating the denominator term:
$$ D = 20(60-20) – 0.2(40)(100-20) = 20(40) – 8(80) = 800 – 640 = 160 $$
Calculating the numerator term inside bracket:
$$ N = 20(100-60) = 20(40) = 800 $$
$$ T_{total} = 40 \left[ 1 + \frac{800}{160} \right] = 40 [1 + 5] = 240\,\text{min} $$
(b) Check for Liquid with Boiling Point $150^\circ\text{C}$
We use the same expression but with $\theta_b’ = 150^\circ\text{C}$. Let’s check the denominator:
$$ D’ = V_i (\theta_1 – \theta_i) – r \Delta t_1 (\theta_b’ – \theta_i) $$
$$ D’ = 20(40) – 8(150 – 20) = 800 – 8(130) = 800 – 1040 = -240 $$
Since the denominator is negative, the time calculated would be negative, which is physically impossible. This indicates that the cooling rate due to the inflow exceeds the heating rate at higher temperatures, preventing the liquid from ever reaching $150^\circ\text{C}$.
The maximum attainable temperature $\theta_{max}$ occurs when the net heating rate is zero (Denominator = 0):
$$ V_i (\theta_1 – \theta_i) = r \Delta t_1 (\theta_{max} – \theta_i) $$
$$ 800 = 8(\theta_{max} – 20) \Rightarrow 100 = \theta_{max} – 20 \Rightarrow \theta_{max} = 120^\circ\text{C} $$
Since $150^\circ\text{C} > 120^\circ\text{C}$, the mixture will never boil.