Question

While camping, some students decide to make hot chocolate by heating water with a solar heater that focuses sunlight onto a small area. Sunlight falls on their solar heater, of area \(1.5 \mathrm{m}^{2},\) with an intensity of $750 \mathrm{W} / \mathrm{m}^{2}$ How long will it take 1.0 L of water at \(15.0^{\circ} \mathrm{C}\) to rise to a boiling temperature of $100.0^{\circ} \mathrm{C} ?$

Short answer

Answer: It will take approximately 316 seconds for 1.0 L of water to heat up from 15.0°C to 100.0°C using the solar heater.

Step-by-step solution

Find the power produced by the solar heater

Given the area of the solar heater (1.5 m²) and the sunlight intensity (750 W/m²), we can find the power (P) produced by the solar heater using the formula: $$ P = \text{area} \times \text{intensity} $$ So, $$ P = 1.5 \mathrm{m}^{2} \times 750 \frac{\mathrm{W}}{\mathrm{m}^{2}} $$ Calculate the power: $$ P = 1125 W $$

Calculate the energy required to heat the water

Next, we will find the energy (E) required to heat 1.0 L of water from 15.0°C to 100.0°C. We know the specific heat capacity (c) of water is \(4186 \frac{\mathrm{J}}{\mathrm{kg} \cdot \mathrm{K}}\), and 1 L of water has a mass (m) of 1 kg. We will use the formula: $$ E = mc\Delta T $$ Where \(\Delta T\) is the change in temperature. For this problem, \(\Delta T = 100.0 - 15.0 = 85.0\degree \mathrm{C} \). So, $$ E = 1 \mathrm{kg} \times 4186 \frac{\mathrm{J}}{\mathrm{kg} \cdot \mathrm{K}} \times 85.0\degree \mathrm{C} $$ Calculate the energy: $$ E = 355310 \mathrm{J} $$

Determine the time it takes for the water to reach 100.0°C

Now that we have the power and energy, we can find the time it takes for the water to reach 100.0°C. We will use the formula: $$ t = \frac{E}{P} $$ Plugging in the values: $$ t = \frac{355310 \mathrm{J}}{1125 \mathrm{W}} $$ Calculate the time: $$ t = 316 \mathrm{s} $$ It will take approximately 316 seconds for 1.0 L of water to heat up from 15.0°C to 100.0°C using the solar heater.