Solution
System Defined: We have a heater plate with heat capacity $H$ and water. The heater supplies constant power $P$.
Let $\theta_b = 100^\circ$C be the boiling point and $\theta_i = 0^\circ$C be the initial temperature.
Phase 1: Heating $m_1$
The heater is on for time $\Delta t_1$. This energy heats the plate (capacity $H$) and the water ($m_1$) from $\theta_i$ to $\theta_b$.
$$ P \Delta t_1 = (H + m_1 s) (\theta_b – \theta_i) \quad \dots (1) $$
Phase 2: Adding $m_2$
When cold water $m_2$ is added, the system cools down. The heater runs for an additional $\Delta t_2$ to bring everything back to boiling.
From an energy conservation perspective, the net energy supplied during $\Delta t_2$ is used solely to raise the added mass $m_2$ from $\theta_i$ to $\theta_b$. (The plate and $m_1$ started at $\theta_b$ and ended at $\theta_b$).
$$ P \Delta t_2 = m_2 s (\theta_b – \theta_i) \quad \dots (2) $$
Determining Heat Capacity of Plate ($H$):
From equation (2), we find the power $P$:
$$ P = \frac{m_2 s (\theta_b – \theta_i)}{\Delta t_2} $$
Substitute this into equation (1):
$$ \left[ \frac{m_2 s (\theta_b – \theta_i)}{\Delta t_2} \right] \Delta t_1 = (H + m_1 s) (\theta_b – \theta_i) $$
Cancel $(\theta_b – \theta_i)$:
$$ m_2 s \frac{\Delta t_1}{\Delta t_2} = H + m_1 s $$
$$ H = s \left( m_2 \frac{\Delta t_1}{\Delta t_2} – m_1 \right) $$
Phase 3: Evaporation
The heater is switched off. The heat stored in the massive plate ($H \theta_b$, assuming relative to 0) is now transferred to the water to cause evaporation (Latent heat $L$).
$$ \text{Heat Stored} = \text{Heat for Evaporation} $$
$$ H (\theta_b – \theta_i) = m_{evap} L $$
$$ m_{evap} = \frac{H (\theta_b – \theta_i)}{L} $$
Substitute $H$:
$$ m_{evap} = \frac{s (\theta_b – \theta_i)}{L} \left( m_2 \frac{\Delta t_1}{\Delta t_2} – m_1 \right) $$
Numerical Substitution:
Given: $m_1 = 1.0 \text{ kg}, m_2 = 1.0 \text{ kg}$.
$\Delta t_1 = 15 \text{ min}, \Delta t_2 = 10 \text{ min}$.
$s = 4.2 \text{ kJ/kg}^\circ\text{C} = 4200 \text{ J/kg}^\circ\text{C}$.
$L = 2.1 \times 10^6 \text{ J/kg}$.
$$ m_{evap} = \frac{4200 (100)}{2.1 \times 10^6} \left( 1.0 \times \frac{15}{10} – 1.0 \right) $$
$$ m_{evap} = \frac{4.2 \times 10^5}{2.1 \times 10^6} \left( 1.5 – 1.0 \right) $$
$$ m_{evap} = 0.2 \times 0.5 = 0.1 \text{ kg} $$
Mass Evaporated = 100 g