THERMAL BYU 11

• Mass of ice, $m_{ice} = 1.0 \text{ kg}$
• Mass of water, $m_{water} = 1.0 \text{ kg}$
• Total initial mass = $2.0 \text{ kg}$
• Base Area, $S = 200 \text{ cm}^2 = 0.02 \text{ m}^2$
• Mass of kerosene burned, $m_k = 34 \text{ g} = 0.034 \text{ kg}$
• Calorific value of kerosene, $q = 7.0 \times 10^4 \text{ kJ/kg} = 7.0 \times 10^7 \text{ J/kg}$
• Efficiency, $\eta = 50\% = 0.5$
• Latent heat of ice, $L_i = 3.5 \times 10^5 \text{ J/kg}$
• Specific heat of water, $s_w = 4200 \text{ J/kg}^\circ\text{C}$
• Latent heat of vaporization, $L_v = 2.5 \times 10^6 \text{ J/kg}$
• Density of water, $\rho = 1000 \text{ kg/m}^3$

Step A: Heat Supplied
The effective heat supplied to the mixture is determined by the kerosene burned and the efficiency of the stove.

$$ Q_{supplied} = \eta \times m_k \times q $$

$$ Q_{supplied} = 0.5 \times 0.034 \text{ kg} \times 7.0 \times 10^7 \text{ J/kg} $$

$$ Q_{supplied} = 1,190,000 \text{ J} = 1.19 \text{ MJ} $$

Step B: Heat to Melt Ice
The mixture is at $0^\circ\text{C}$. First, the $1.0 \text{ kg}$ of ice must melt.

$$ Q_{melt} = m_{ice} \times L_i = 1.0 \times 3.5 \times 10^5 = 350,000 \text{ J} = 0.35 \text{ MJ} $$

Remaining Heat: $Q_{rem1} = 1.19 – 0.35 = 0.84 \text{ MJ}$.
Since $Q_{rem1} > 0$, all ice melts. We now have $2.0 \text{ kg}$ of water at $0^\circ\text{C}$.

Step C: Heat to Boil Water
Now we heat $2.0 \text{ kg}$ of water from $0^\circ\text{C}$ to $100^\circ\text{C}$.

$$ Q_{heat} = m_{total} \times s_w \times \Delta T $$

$$ Q_{heat} = 2.0 \times 4200 \times 100 = 840,000 \text{ J} = 0.84 \text{ MJ} $$

Remaining Heat: $Q_{rem2} = Q_{rem1} – Q_{heat} = 0.84 – 0.84 = 0 \text{ J}$.

Since the remaining heat energy is exactly zero after raising the water temperature to $100^\circ\text{C}$, no water is converted into steam. The total mass of water remaining is exactly the initial total mass.

$$ m_{final} = 2.0 \text{ kg} $$

To find the height ($h$) of the water level:

$$ \text{Volume } V = \frac{m_{final}}{\rho} = \frac{2.0 \text{ kg}}{1000 \text{ kg/m}^3} = 2.0 \times 10^{-3} \text{ m}^3 $$

$$ h = \frac{V}{S} = \frac{2.0 \times 10^{-3} \text{ m}^3}{0.02 \text{ m}^2} $$

$$ h = 0.1 \text{ m} = 10 \text{ cm} $$