1. **Stating the problem:** A liquid is heated at a constant rate. The temperature rises by 0.50 °C per second until boiling. After boiling, in 10 minutes, half the mass vaporizes. We need to find: - Specific heat capacity $c$ of the liquid - Specific latent heat of vaporization $L$ 2. **Given data:** - Temperature rise rate: $0.50\ \degree C/s$ - Time to boil: unknown, but heating until boiling - After boiling, time $t = 10$ minutes $= 600$ seconds - Half the mass vaporizes in 10 minutes 3. **Formulas:** - Heat to raise temperature: $Q_1 = mc\Delta T$ - Power supplied: $P = \frac{Q_1}{t_1} = mc\frac{\Delta T}{t_1}$ where $t_1$ is time to reach boiling - Heat to vaporize: $Q_2 = mL$ - Power supplied during vaporization: $P = \frac{Q_2}{t_2} = \frac{mL}{t_2}$ where $t_2 = 600$ s 4. **Step 1: Calculate time to reach boiling $t_1$** Assuming temperature rises at $0.50\ \degree C/s$ until boiling point $\Delta T$ (unknown), $$ t_1 = \frac{\Delta T}{0.50} $$ 5. **Step 2: Express power $P$ during heating:** $$ P = mc \frac{\Delta T}{t_1} = mc \frac{\Delta T}{\frac{\Delta T}{0.50}} = mc \times 0.50 $$ So, $$ P = 0.50 mc $$ 6. **Step 3: Express power $P$ during vaporization:** Half the mass vaporizes in $t_2 = 600$ s, so heat required: $$ Q_2 = \frac{m}{2} L $$ Power during vaporization: $$ P = \frac{Q_2}{t_2} = \frac{\frac{m}{2} L}{600} = \frac{mL}{1200} $$ 7. **Step 4: Equate power during heating and vaporization (constant rate):** $$ 0.50 mc = \frac{mL}{1200} $$ Cancel $m$: $$ 0.50 c = \frac{L}{1200} $$ Multiply both sides by 1200: $$ 1200 \times 0.50 c = L $$ $$ 600 c = L $$ 8. **Step 5: Use options to find $c$ and $L$:** Options for $c$ are fractions: $\frac{1}{600}, \frac{1}{300}, \frac{1}{10}, \frac{1}{5}$ Calculate $L$ for each: - If $c = \frac{1}{600}$, then $L = 600 \times \frac{1}{600} = 1$ - If $c = \frac{1}{300}$, then $L = 600 \times \frac{1}{300} = 2$ - If $c = \frac{1}{10}$, then $L = 600 \times \frac{1}{10} = 60$ - If $c = \frac{1}{5}$, then $L = 600 \times \frac{1}{5} = 120$ Since latent heat is usually much larger than specific heat capacity, the reasonable choice is: $$ c = \frac{1}{600} $$ $$ L = 1 $$ 9. **Final answers:** - Specific heat capacity $c = \frac{1}{600}$ - Specific latent heat of vaporization $L = 1$ (in consistent units)