/*! 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 78 On a cold day, 24500 J of heat l... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

On a cold day, 24500 J of heat leaks out of a house. The inside temperature is \(21^{\circ} \mathrm{C},\) and the outside temperature is \(-15^{\circ} \mathrm{C} .\) What is the increase in the entropy of the universe that this heat loss produces?

Short Answer

Expert verified
The increase in the entropy of the universe is 11.63 J/K.

Step by step solution

01

Identify the Given Values

The heat lost by the house is given as 24500 J. The inside temperature is \( T_{inside} = 21^{\circ}C \) which can be converted to Kelvin as \( 21 + 273.15 = 294.15 K \). The outside temperature is \( T_{outside} = -15^{\circ}C \) which can be converted to Kelvin as \( -15 + 273.15 = 258.15 K \).
02

Calculate Entropy Change Inside the House

The entropy change \( \Delta S_{inside} \) when heat \( Q \) leaves the house is calculated using the formula \( \Delta S_{inside} = -\frac{Q}{T_{inside}} \). Substituting the values, we have \( \Delta S_{inside} = -\frac{24500 \ J}{294.15 \ K} \approx -83.29 \ J/K \).
03

Calculate Entropy Change Outside the House

The entropy change \( \Delta S_{outside} \) when the same heat \( Q \) is absorbed by the surroundings (outside the house) is \( \Delta S_{outside} = \frac{Q}{T_{outside}} \). Substituting the values, we have \( \Delta S_{outside} = \frac{24500 \ J}{258.15 \ K} \approx 94.92 \ J/K \).
04

Calculate Total Entropy Change

The total change in entropy of the universe is the sum of the entropy changes inside and outside the house: \( \Delta S_{universe} = \Delta S_{inside} + \Delta S_{outside} \). Substituting the calculated values, \( \Delta S_{universe} = -83.29 \ J/K + 94.92 \ J/K = 11.63 \ J/K \).
05

Conclude the Entropy Change

The increase in the entropy of the universe, resulting from the heat loss from the house to the outside, is \( 11.63 \ J/K \).

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 91Ó°ÊÓ!

Key Concepts

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

Heat Transfer
Heat transfer is the process by which thermal energy moves from one place to another due to temperature differences. Understanding heat transfer is essential for solving many thermodynamic problems like the one in this exercise.

In our example, heat is leaking from inside the house to the outside environment. This movement of heat occurs because of a temperature gradient; heat naturally flows from the warmer area (inside the house) to the cooler one (outside). This kind of heat transfer is driven by conduction, which is the transfer of thermal energy through a material without the movement of the material itself.

To calculate changes due to heat transfer, temperatures must be measured in Kelvin. Celsius temperatures are converted to Kelvin by adding 273.15. This conversion is essential as Kelvin is the standard unit for temperature in equations involving heat transfer and thermodynamics, ensuring consistency and accuracy in calculations.
Thermodynamics
Thermodynamics is the scientific study of heat, work, and the associated energy transfers. It's a fundamental principle that describes the behavior of energy within physical systems.

One of the core concepts of thermodynamics is the First Law, which is essentially the law of conservation of energy. It states that energy cannot be created or destroyed, only transformed from one form to another. In our example from the exercise, this principle is illustrated through the manner in which heat energy moves from the inside to the outside of the house.

The Second Law of Thermodynamics, relating to entropy, indicates that energy systems naturally progress towards a state of disorder or randomness. Therefore, when heat leaves the house, the overall entropy—or disorder—increases in the universe. This idea of movement towards greater entropy underlies why total entropy increases even though the house loses energy.
Entropy Change
Entropy is a measure of disorder or randomness in a system. When dealing with the entropy change due to heat transfer, this concept becomes crucial. The main idea is that when heat is transferred, the entropy of the system and its surroundings changes.

In the given exercise, two entropy changes occur: one inside the house and one outside. Inside the house, as heat leaves, the entropy decreases, which is calculated as \( \Delta S_{inside} = -\frac{Q}{T_{inside}} \). Outside, where the heat is absorbed, the entropy increases, calculated by \( \Delta S_{outside} = \frac{Q}{T_{outside}} \).

The total change in entropy of the universe is determined by adding these two changes: \( \Delta S_{universe} = \Delta S_{inside} + \Delta S_{outside} \). In this way, we see that even though entropy decreases within a closed space, the increase in the environment results in a net increase in the universe's entropy. This conclusion aligns with the Second Law of Thermodynamics, confirming that the total entropy, that is, the overall disorder, tends to increase.

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

Three moles of a monatomic ideal gas are heated at a constant volume of \(1.50 \mathrm{m}^{3} .\) The amount of heat added is \(5.24 \times 10^{3} \mathrm{J} .\) (a) What is the change in the temperature of the gas? (b) Find the change in its internal energy. (c) Determine the change in pressure.

The sun is a sphere with a radius of \(6.96 \times 10^{8} \mathrm{m}\) and an average surface temperature of \(5800 \mathrm{K}\). Determine the amount by which the sun's thermal radiation increases the entropy of the entire universe each second. Assume that the sun is a perfect blackbody, and that the average temperature of the rest of the universe is \(2.73 \mathrm{K}\). Do not consider the thermal radiation absorbed by the sun from the rest of the universe.

The water in a deep underground well is used as the cold reservoir of a Carnot heat pump that maintains the temperature of a house at \(301 \mathrm{K}\). To deposit \(14200 \mathrm{J}\) of heat in the house, the heat pump requires \(800 \mathrm{J}\) of work. Determine the temperature of the well water.

A system does \(4.8 \times 10^{4} \mathrm{J}\) of work, and \(7.6 \times 10^{4} \mathrm{J}\) of heat flows into the system during the process. Find the change in the internal energy of the system.

A Carnot engine has an efficiency of 0.700 , and the temperature of its cold reservoir is \(378 \mathrm{K}\). (a) Determine the temperature of its hot reservoir. (b) If 5230 J of heat is rejected to the cold reservoir, what amount of heat is put into the engine?

See all solutions

Recommended explanations on Physics Textbooks

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.