/*! 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 29 In a solar water heater, energy ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 18: Problem 29

In a solar water heater, energy from the Sun is gathered by water that circulates through tubes in a rooftop collector. The solar radiation enters the collector through a transparent cover and warms the water in the tubes; this water is pumped into a holding tank. Assume that the efficiency of the overall system is \(20 \%\) (that is, \(80 \%\) of the incident solar energy is lost from the system). What collector area is necessary to raise the temperature of 200 \(\mathrm{L}\) of water in the tank from \(20^{\circ} \mathrm{C}\) to \(40^{\circ} \mathrm{C}\) in \(1.0 \mathrm{~h}\) when the intensity of incident sunlight is \(700 \mathrm{~W} / \mathrm{m}^{2}\) ?

Short Answer

Expert verified
The necessary collector area is approximately 33.22 m虏.

Step by step solution

01

Calculate the Energy Needed to Heat the Water

To find out how much energy is needed to heat the water, we use the formula for heat energy: \( Q = m \cdot c \cdot \Delta T \). Here, \( m \) is the mass of the water, \( c \) is the specific heat capacity of water (which is approximately \( 4.186 \text{ J/g掳C} \)), and \( \Delta T \) is the change in temperature.First, convert the volume of water to mass. Since the density of water is \( 1 \text{ kg/L} \), 200 L corresponds to 200 kg.The change in temperature \( \Delta T \) is \(40^{\circ}\text{C} - 20^{\circ}\text{C} = 20^{\circ}\text{C}\).Plugging these values in:\[Q = 200 \times 4,186 \times 20 = 16,744,000 \text{ J}\]
02

Determine Solar Energy Required by Accounting for Efficiency Loss

With an efficiency of 20%, only 20% of the solar energy collected is used to heat the water. Therefore, the actual solar energy required is:\[\frac{16,744,000}{0.20} = 83,720,000 \text{ J}\]
03

Calculate the Power Collected by the Collectors

The power collected by the collectors is the intensity of sunlight times the collector area, integrated over time. With the intensity given as \(700 \text{ W/m}^2\), over one hour (3600 seconds), the energy collected is:\[ ext{Energy Collected} = 700 \times A \times 3600\]where \( A \) is the area of the collectors.
04

Solve for the Collector Area

Set the energy collected equal to the energy required:\[700 \times A \times 3600 = 83,720,000 \2520 A = 83,720,000 \A = \frac{83,720,000}{2520} \approx 33,222.22 \text{ m}^2\]
05

Concluding the Calculation

Therefore, the collector area required to raise the temperature of the water is approximately \(33.22 \text{ m}^2\).

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!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Energy Conversion Efficiency
Energy conversion efficiency is a crucial concept in understanding how systems like solar water heaters function. It measures the ratio of useful energy output to the energy input from the sun that is actually utilized, often expressed as a percentage. In our exercise, the solar water heater has an efficiency of 20%. This means that only 20% of the incident solar energy is effectively used to heat the water, while the remaining 80% is lost.
Understanding efficiency helps us gauge how well the water heater performs and how much energy we need to collect to meet our heating requirements. It necessitates at times a larger collector area or advanced technology to minimize energy loss. Efficiency considerations are pivotal when designing systems to ensure sustainable and optimal energy usage.
Collector Area Calculation
Calculating the collector area for a solar water heater involves figuring out how much surface area is necessary to collect sufficient solar energy for heating water. Given an energy requirement, which accounts for both the specification of the energy demand and the system efficiency, we can calculate the area needed.
The formula used is based on the relation:
  • Energy collected = Solar intensity 脳 Collector area 脳 Time of exposure
  • Energy requirement (adjusted for efficiency) = Solar energy necessary to heat the water
In this context, if we have an intensity of 700 W/m虏, and a need to raise 200 L of water by a certain temperature change, we compute:\[700 \times A \times 3600 = 83,720,000 \text{ J}\]Through rearranging and solving for the collector area, we derive that:\[A = \frac{83,720,000}{2520} \approx 33.22 \text{ m}^2\]This indicates how much area is necessary to achieve the desired temperature rise, assuming all factors remain constant.
Specific Heat Capacity of Water
The specific heat capacity of water plays an essential role in calculating the energy required to heat it. The specific heat capacity is denoted as \(c\) and for water, it is approximately \(4.186 \, \text{J/g掳C}\). This value indicates the amount of energy needed to raise the temperature of 1 gram of water by 1 degree Celsius.
To calculate the energy to heat a given quantity of water, the formula utilized is:\[Q = m \times c \times \Delta T\]where:
  • \(m\) is the mass of the water, converted from volume under the physical property that the density of water is 1 kg/L
  • \(c\) is the specific heat capacity of water
  • \(\Delta T\) is the temperature change required
In our exercise, to increase the temperature of 200 L of water from 20掳C to 40掳C, we compute:
\[Q = 200 \times 4,186 \times 20 = 16,744,000 \text{ J}\]This mathematical application helps illustrate how specific heat capacity is a determining factor in energy calculations for temperature adjustments.

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 \(1700 \mathrm{~kg}\) Buick moving at \(83 \mathrm{~km} / \mathrm{h}\) brakes to a stop, at uniform deceleration and without skidding, over a distance of \(93 \mathrm{~m}\). At what average rate is mechanical energy transferred to thermal energy in the brake system?

To withstand the harsh weather of the Antarctic, emperor penguins huddle in groups (Fig. \(18-50)\). Assume that a penguin is a circular cylinder with a top surface area \(a=0.34 \mathrm{~m}^{2}\) and height \(h=1.1 \mathrm{~m}\). Let \(P_{r}\) be the rate at which an individual penguin radiates energy to the environment (through the top and the sides); thus \(N P_{r}\) is the rate at which \(N\) identical, wellseparated penguins radiate. If the penguins huddle closely to form a huddled cylinder with top surface area \(N a\) and height \(h\), the cylinder radiates at the rate \(P_{h}\). If \(N=1000\), (a) what is the value of the fraction \(P_{h} / N P_{r}\) and (b) by what percentage does huddling reduce the total radiation loss?

On finding your stove out of order, you decide to boil the water for a cup of tea by shaking it in a thermos flask. Suppose that you use tap water at \(19^{\circ} \mathrm{C}\), the water falls \(32 \mathrm{~cm}\) each shake, and you make 27 shakes each minute. Neglecting any loss of thermal energy by the flask, how long (in minutes) must you shake the flask until the water reaches \(100^{\circ} \mathrm{C}\) ?

Liquid water coats an active (growing) icicle and extends up a short, narrow tube along the central axis (Fig. \(18-55)\). Because the water-ice interface must have a temperature of \(0{ }^{\circ} \mathrm{C}\), the water in the tube cannot lose energy through the sides of the icicle or down through the tip because there is no temperature change in those directions. It can lose energy and freeze only by sending energy up (through distance \(L\) ) to the top of the icicle, where the temperature \(T_{r}\) can be below \(0^{\circ} \mathrm{C}\). Take \(L=0.12 \mathrm{~m}\) and \(T_{r}=-5^{\circ} \mathrm{C}\). Assume that the central tube and the upward conduction path both have cross-sectional area \(A .\) In terms of \(A\), what rate is (a) energy conducted upward and (b) mass converted from liquid to ice at the top of the central tube? (c) At what rate does the top of the tube move downward because of water freezing there? The thermal conductivity of ice is \(0.400 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), and the density of liquid water is \(1000 \mathrm{~kg} / \mathrm{m}^{3}\).

A certain diet doctor encourages people to diet by drinking ice water. His theory is that the body must burn off enough fat to raise the temperature of the water from \(0.00^{\circ} \mathrm{C}\) to the body temperature of \(37.0^{\circ} \mathrm{C}\). How many liters of ice water would have to be consumed to burn off \(454 \mathrm{~g}\) (about \(1 \mathrm{lb}\) ) of fat, assuming that burning this much fat requires 3500 Cal be transferred to the ice water? Why is it not advisable to follow this diet? (One liter \(=10^{3} \mathrm{~cm}^{3}\). The density of water is \(1.00 \mathrm{~g} / \mathrm{cm}^{3}\).)
See all solutions

Recommended explanations on Physics Textbooks

Force

Read Explanation

Quantum Physics

Read Explanation

Physical Quantities And Units

Read Explanation

Linear Momentum

Read Explanation

Electrostatics

Read Explanation

Rotational Dynamics

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.