/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 73 While camping, some students dec... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

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掳C to 100.0掳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 m虏) and the sunlight intensity (750 W/m虏), 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掳C to 100.0掳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掳C

Now that we have the power and energy, we can find the time it takes for the water to reach 100.0掳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掳C to 100.0掳C using the solar heater.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A high jumper of mass \(60.0 \mathrm{kg}\) consumes a meal of $3.00 \times 10^{3} \mathrm{kcal}\( prior to a jump. If \)3.3 \%$ of the energy from the food could be converted to gravitational potential energy in a single jump, how high could the athlete jump?
A cylinder contains 250 L of hydrogen gas \(\left(\mathrm{H}_{2}\right)\) at \(0.0^{\circ} \mathrm{C}\) and a pressure of 10.0 atm. How much energy is required to raise the temperature of this gas to \(25.0^{\circ} \mathrm{C} ?\)
A thermometer containing \(0.10 \mathrm{g}\) of mercury is cooled from \(15.0^{\circ} \mathrm{C}\) to \(8.5^{\circ} \mathrm{C} .\) How much energy left the mercury in this process?
Five ice cubes, each with a mass of \(22.0 \mathrm{g}\) and at a temperature of \(-50.0^{\circ} \mathrm{C},\) are placed in an insulating container. How much heat will it take to change the ice cubes completely into steam?

Consider the leaf of Problem \(70 .\) Assume that the top surface of the leaf absorbs \(70.0 \%\) of \(9.00 \times 10^{2} \mathrm{W} / \mathrm{m}^{2}\) of radiant energy, while the bottom surface absorbs all of the radiant energy incident on it due to its surroundings at \(25.0^{\circ} \mathrm{C} .\) (a) If the only method of heat loss for the leaf were thermal radiation, what would be the temperature of the leaf? (Assume that the leaf radiates like a blackbody.) (b) If the leaf is to remain at a temperature of \(25.0^{\circ} \mathrm{C}\) how much power per unit area must be lost by other methods such as transpiration (evaporative heat loss)?
See all solutions

Recommended explanations on Physics Textbooks

Energy Physics

Read Explanation

Solid State Physics

Read Explanation

Electromagnetism

Read Explanation

Electricity and Magnetism

Read Explanation

Fields in Physics

Read Explanation

Medical Physics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.