Log out Go to App
Go to App

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°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.

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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

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 tungsten filament in a lamp is heated to a temperature of $2.6 \times 10^{3} \mathrm{K}\( by an electric current. The tungsten has an emissivity of \)0.32 .$ What is the surface area of the filament if the lamp delivers $40.0 \mathrm{W}$ of power?
An \(83-\mathrm{kg}\) man eats a banana of energy content $1.00 \times 10^{2} \mathrm{kcal} .$ If all of the energy from the banana is converted into kinetic energy of the man, how fast is he moving, assuming he starts from rest?
A 20.0 -g lead bullet leaves a rifle at a temperature of $47.0^{\circ} \mathrm{C}\( and travels at a velocity of \)5.00 \times 10^{2} \mathrm{m} / \mathrm{s}\( until it hits a large block of ice at \)0^{\circ} \mathrm{C}$ and comes to rest within it. How much ice will melt?
A copper bar of thermal conductivity $401 \mathrm{W} /(\mathrm{m} \cdot \mathrm{K})\( has one end at \)104^{\circ} \mathrm{C}$ and the other end at \(24^{\circ} \mathrm{C}\). The length of the bar is \(0.10 \mathrm{m}\) and the cross-sectional area is \(1.0 \times 10^{-6} \mathrm{m}^{2} .\) (a) What is the rate of heat conduction, \(\mathscr{P}\) along the bar? (b) What is the temperature gradient in the bar? (c) If two such bars were placed in series (end to end) between the same temperature baths, what would \(\mathscr{P}\) be? (d) If two such bars were placed in parallel (side by side) with the ends in the same temperature baths, what would \(\mathscr{P}\) be? (e) In the series case, what is the temperature at the junction where the bars meet?
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?
See all solutions

Recommended explanations on Physics Textbooks

Modern Physics

Read Explanation

Dynamics

Read Explanation

Classical Mechanics

Read Explanation

Force

Read Explanation

Waves Physics

Read Explanation

Electricity

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.

Sign-up for free