Solution 13: Calorimetry and Heating
Given:
- Initial state: Equal mass of ice ($m$) and water ($m$). Total mass = $2m$.
- Melting phase: All ice melts in $\Delta t_0 = 160$ min.
- Heating phase: Time ($t$) required to heat the water (mass $2m$) from $\theta_1 = 22^\circ \text{C}$ to $\theta_2 = 23^\circ \text{C}$.
Step 1: Melting Process
The rate of heat gain follows Newton’s Law of Cooling/Heating: $\frac{dQ}{dt} = k(\theta_r – \theta_{sys})$. During melting, the system temperature is constant at $0^\circ \text{C}$. $$Q_{melt} = mL_f$$ Rate of heat flow: $\frac{dQ}{dt} \approx \frac{mL}{\Delta t_0}$ $$ \frac{mL}{\Delta t_0} = k(25 – 0) \Rightarrow mL = 25k(160) \quad \dots \text{(Eq. 1)}$$
The rate of heat gain follows Newton’s Law of Cooling/Heating: $\frac{dQ}{dt} = k(\theta_r – \theta_{sys})$. During melting, the system temperature is constant at $0^\circ \text{C}$. $$Q_{melt} = mL_f$$ Rate of heat flow: $\frac{dQ}{dt} \approx \frac{mL}{\Delta t_0}$ $$ \frac{mL}{\Delta t_0} = k(25 – 0) \Rightarrow mL = 25k(160) \quad \dots \text{(Eq. 1)}$$
Step 2: Heating Process
After the ice melts, the total mass is $2m$. We heat this from $22^\circ \text{C}$ to $23^\circ \text{C}$. Since the temperature interval is small, we use the average temperature for the rate calculation. $$\theta_{avg} = \frac{22 + 23}{2} = 22.5^\circ \text{C}$$ Temperature difference driving heat flow: $\Delta \theta = 25 – 22.5 = 2.5^\circ \text{C}$. Heat required: $$Q_{heat} = (2m)s_w(\theta_2 – \theta_1) = 2ms_w(1)$$ Rate of heat flow at this instant: $$\frac{dQ}{dt} = k(25 – 22.5) = 2.5k$$ Time taken ($t$): $$t = \frac{\text{Heat Required}}{\text{Rate}} = \frac{2ms_w}{2.5k}$$
After the ice melts, the total mass is $2m$. We heat this from $22^\circ \text{C}$ to $23^\circ \text{C}$. Since the temperature interval is small, we use the average temperature for the rate calculation. $$\theta_{avg} = \frac{22 + 23}{2} = 22.5^\circ \text{C}$$ Temperature difference driving heat flow: $\Delta \theta = 25 – 22.5 = 2.5^\circ \text{C}$. Heat required: $$Q_{heat} = (2m)s_w(\theta_2 – \theta_1) = 2ms_w(1)$$ Rate of heat flow at this instant: $$\frac{dQ}{dt} = k(25 – 22.5) = 2.5k$$ Time taken ($t$): $$t = \frac{\text{Heat Required}}{\text{Rate}} = \frac{2ms_w}{2.5k}$$
Step 3: Solving for t
From Eq. 1, $k = \frac{mL}{25 \times 160}$. Substitute this into the time equation: $$t = \frac{2ms_w}{2.5 \left( \frac{mL}{25 \times 160} \right)}$$ Mass $m$ cancels out: $$t = \frac{2s_w \times 25 \times 160}{2.5 \times L}$$ $$t = \frac{2(4.2) \times 4000}{2.5 \times 320}$$ $$t = \frac{33600}{800} = 42 \text{ min}$$
From Eq. 1, $k = \frac{mL}{25 \times 160}$. Substitute this into the time equation: $$t = \frac{2ms_w}{2.5 \left( \frac{mL}{25 \times 160} \right)}$$ Mass $m$ cancels out: $$t = \frac{2s_w \times 25 \times 160}{2.5 \times L}$$ $$t = \frac{2(4.2) \times 4000}{2.5 \times 320}$$ $$t = \frac{33600}{800} = 42 \text{ min}$$
Answer: It would take 42 minutes.