/*! 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 70 70 In a certain solar house, ene... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 18: Problem 70

70 In a certain solar house, energy from the Sun is stored in barrels filled with water. In a particular winter stretch of five cloudy days, \(1.00 \times 10^{6} \mathrm{kcal}\) is needed to maintain the inside of the house at \(22.0^{\circ} \mathrm{C}\). Assuming that the water in the barrels is at \(50.0^{\circ} \mathrm{C}\) and that the water has a density of \(1.00 \times 10^{3} \mathrm{~kg} / \mathrm{m}^{3}\), what volume of water is required?

Short Answer

Expert verified
The volume of water required is approximately 35.7 m鲁.

Step by step solution

01

Understanding the Problem

We need to find the volume of water needed to provide a specific amount of energy to maintain the temperature of a house. The given energy requirement is \(1.00 \times 10^6 \text{kcal}\). We know the initial temperature of the water is \(50.0^{\circ} \mathrm{C}\) and the final temperature needed is \(22.0^{\circ} \mathrm{C}\). The task is to determine how much water, in terms of volume, is needed for this energy transfer.
02

Convert Energy Requirement to Joules

The energy needed is given in kcal. Convert this to joules using the conversion factor: \(1 \text{kcal} = 4184 \text{J}\). So, \(1.00 \times 10^6 \text{kcal} = 1.00 \times 10^6 \times 4184 \text{J} = 4.184 \times 10^9 \text{J}\).
03

Calculate Heat Transfer per Unit Mass

The heat transferred due to cooling of water can be calculated using the formula: \[Q = mc\Delta T\]where \(m\) is mass, \(c\) is the specific heat capacity (for water \(c = 4186 \text{ J/kg掳C}\)), and \(\Delta T\) is the temperature change. Here, \(\Delta T = 50.0^{\circ}C - 22.0^{\circ}C = 28.0^{\circ}C\).
04

Determine Mass of Water Needed

From \(Q = mc\Delta T\), the mass can be obtained as:\[m = \frac{Q}{c\Delta T} = \frac{4.184 \times 10^9}{4186 \times 28}\]Calculate the mass: \[m \approx 35,706 \text{ kg}\]
05

Convert Mass to Volume

With the density \(\rho = 1000 \text{ kg/m}^3\) and using the mass from the previous step:\[V = \frac{m}{\rho} = \frac{35,706}{1000} \approx 35.706 \text{ m}^3\]
06

Conclusion of Volume Required

The volume of water required to meet the energy needs for the five days is approximately \(35.7 \text{ m}^3\).

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.

Heat Transfer
Heat transfer is the process of energy moving from one place or material to another due to a temperature difference. In our exercise, the energy from the solar house's water barrels is transferred to keep the house warm. This involves thermal energy moving from the hot water (initially at 50.0掳C) to the cooler environment (maintained at 22.0掳C).

Key concepts to understand here include:
  • Temperature difference: The temperature change (\[\Delta T\]) encourages heat transfer. In our context, this is a drop from 50掳C to 22掳C, which translates to a 28掳C change.
  • Direction of heat flow: Heat naturally flows from hotter substances to cooler ones until thermal equilibrium is reached.
  • Thermal equilibrium: A state where no net heat transfer occurs between objects because they are at the same temperature.
These concepts underline the basis of calculating heat transfer in thermodynamic questions, allowing us to find out how much energy needs to be transferred to achieve the desired temperature change.
Specific Heat Capacity
Specific heat capacity is a crucial property in thermodynamics, describing how much energy is required to raise the temperature of a unit mass of a substance by one degree Celsius. Water, a commonly used medium for temperature regulation because of its high specific heat capacity, can absorb lots of heat with little temperature change.

In mathematical terms, specific heat capacity (\[c\]) is applied in the following formula:\[Q = mc\Delta T\]where:
  • \(Q\) is the heat added or removed
  • \(m\) is the mass of the substance
  • \(c\) is the specific heat capacity (for water, typically 4186 J/kg掳C)
  • \(\Delta T\) is the change in temperature
In this exercise, the specific heat capacity of water allows us to calculate how efficiently water can store and release thermal energy, making it ideal for regulating heat over time in the house. It's this capacity that lets us compute exactly how much water is necessary to release the required energy to maintain a stable house temperature over the cloudy days.
Energy Conversion
Energy conversion refers to the change of energy from one form to another. In the context of the solar house problem, this involves converting thermal energy stored in hot water barrels to useful heat energy that warms the house.

For this task, the process requires understanding the energy conversion that happens in water:
  • Initial energy form: The water is heated by solar energy during sunny days and holds this energy as thermal energy.
  • Conversion process: On cloudy days, this stored thermal energy is released into the house, replacing the need for additional energy sources to maintain warmth.
  • Joule conversion: The specific calculation involves converting the energy given in kilocalories to joules (as energy in physics is usually measured in joules), using the conversion factor \(1 \, \text{kcal} = 4184 \, \text{J}\).
This conversion is vital not only in supporting sustainable energy practices but also offers a practical application of thermodynamics, allowing efficient usage of naturally stored energy to sustain temperatures within the solar house.

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 sphere of radius \(0.500 \mathrm{~m}\), temperature \(27.0^{\circ} \mathrm{C}\), and emissivity \(0.850\) is located in an environment of temperature \(77.0^{\circ} \mathrm{C}\). At what rate does the sphere (a) emit and (b) absorb thermal radiation? (c) What is the sphere's net rate of energy exchange?

Icebergs in the North Atlantic present hazards to shipping, causing the lengths of shipping routes to be increased by about \(30 \%\) during the iceberg season. Attempts to destroy icebergs include planting explosives, bombing, torpedoing, shelling, ramming, and coating with black soot. Suppose that direct melting of the iceberg, by placing heat sources in the ice, is tried. How much energy as heat is required to melt \(10 \%\) of an iceberg that has a mass of 200000 metric tons? (Use 1 metric ton \(=1000 \mathrm{~kg}\).)

Leidenfrost effect. \(\mathrm{A}\) water drop will last about \(1 \mathrm{~s}\) on a hot skillet with a temperature between \(100^{\circ} \mathrm{C}\) and about \(200^{\circ} \mathrm{C}\). However, if the skillet is much hotter, the drop can last several minutes, an effect named after an early investigator. The longer lifetime is due to the support of a thin layer of air and water vapor that separates the drop from the metal (by distance \(L\) in Fig. \(18-48)\). Let \(L=\) \(0.100 \mathrm{~mm}\), and assume that the drop is flat with height \(h=1.50 \mathrm{~mm}\) and bottom face area \(A=4.00 \times 10^{-6} \mathrm{~m}^{2}\). Also assume that the skillet has a constant temperature \(T_{s}=300^{\circ} \mathrm{C}\) and the drop has a temperature of \(100^{\circ} \mathrm{C}\). Water has density \(\rho=1000 \mathrm{~kg} / \mathrm{m}^{3}\), and the supporting layer has thermal conductivity \(k=0.026 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K} .\) (a) At what rate is energy conducted from the skillet to the drop through the drop's bottom surface? (b) If conduction is the primary way energy moves from the skillet to the drop, how long will the drop last?

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 \(700 \mathrm{~W} / \mathrm{m}^{2}\) ?

Suppose that on a linear temperature scale \(X\), water boils at \(-53.5^{\circ} \mathrm{X}\) and freezes at \(-170^{\circ} \mathrm{X}\). What is a temperature of \(340 \mathrm{~K}\) on the \(\mathrm{X}\) scale? (Approximate water's boiling point as \(373 \mathrm{~K}\).)
See all solutions

Recommended explanations on Physics Textbooks

Particle Model of Matter

Read Explanation

Energy Physics

Read Explanation

Fields in Physics

Read Explanation

Wave Optics

Read Explanation

Kinematics Physics

Read Explanation

Thermodynamics

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.