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Chapter 14: Problem 73

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

Expert verified
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

01

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: $$ P = 1125 W $$
02

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: $$ E = 355310 \mathrm{J} $$
03

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.

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Most popular questions from this chapter

A black wood stove has a surface area of \(1.20 \mathrm{m}^{2}\) and a surface temperature of \(175^{\circ} \mathrm{C} .\) What is the net rate at which heat is radiated into the room? The room temperature is \(20^{\circ} \mathrm{C}\).
A person of surface area \(1.80 \mathrm{m}^{2}\) is lying out in the sunlight to get a tan. If the intensity of the incident sunlight is $7.00 \times 10^{2} \mathrm{W} / \mathrm{m}^{2},$ at what rate must heat be lost by the person in order to maintain a constant body temperature? (Assume the effective area of skin exposed to the Sun is \(42 \%\) of the total surface area, \(57 \%\) of the incident radiation is absorbed, and that internal metabolic processes contribute another \(90 \mathrm{W}\) for an inactive person.)
A 10.0 -g iron bullet with a speed of $4.00 \times 10^{2} \mathrm{m} / \mathrm{s}\( and a temperature of \)20.0^{\circ} \mathrm{C}$ is stopped in a \(0.500-\mathrm{kg}\) block of wood, also at \(20.0^{\circ} \mathrm{C} .\) (a) At first all of the bullet's kinetic energy goes into the internal energy of the bullet. Calculate the temperature increase of the bullet. (b) After a short time the bullet and the block come to the same temperature \(T\). Calculate \(T\), assuming no heat is lost to the environment.
If \(125.6 \mathrm{kJ}\) of heat are supplied to \(5.00 \times 10^{2} \mathrm{g}\) of water at \(22^{\circ} \mathrm{C},\) what is the final temperature of the water?
At a tea party, a coffeepot and a teapot are placed on the serving table. The coffeepot is a shiny silver-plated pot with emissivity of \(0.12 ;\) the teapot is ceramic and has an emissivity of \(0.65 .\) Both pots hold \(1.00 \mathrm{L}\) of liquid at \(98^{\circ} \mathrm{C}\) when the party begins. If the room temperature is at \(25^{\circ} \mathrm{C},\) what is the rate of radiative heat loss from the two pots? [Hint: To find the surface area, approximate the pots with cubes of similar volume.]
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