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On a cold winter's day heat leaks slowly out of a house at the rate of \(20.0 \mathrm{kW}\). If the inside temperature is \(22^{\circ} \mathrm{C}\), and the outside temperature is \(-14.5^{\circ} \mathrm{C}\), find the rate of entropy increase.

Short Answer

Expert verified
The rate of entropy increase is approximately 9.56 J/K/s.

Step by step solution

01

Understand the Problem

We need to calculate the rate of entropy increase as heat leaks out of a house. The problem provides us with the rate at which heat is leaking, the inside temperature of the house, and the outside temperature.
02

Determine Heat Loss

The rate of heat loss from the house is given as 20.0 kW, which is equivalent to a heat loss of 20,000 J/s.
03

Convert Temperatures to Kelvin

Entropy calculations require temperatures in Kelvin. To convert Celsius to Kelvin, add 273.15.- Inside temperature in Kelvin: \[ T_{ ext{inside}} = 22 + 273.15 = 295.15 ext{ K} \]- Outside temperature in Kelvin: \[ T_{ ext{outside}} = -14.5 + 273.15 = 258.65 ext{ K} \]
04

Calculate Entropy Change at Each Temperature

The rate of entropy change is calculated using the formula: \[ \Delta S = \frac{Q}{T} \]Where \( Q \) is the rate of heat transfer and \( T \) is the temperature.- Entropy change inside: \[ \Delta S_{ ext{inside}} = \frac{20,000}{295.15} \approx 67.75 ext{ J/K/s} \]- Entropy change outside: \[ \Delta S_{ ext{outside}} = \frac{20,000}{258.65} \approx 77.31 ext{ J/K/s} \]
05

Determine Net Entropy Change

The net rate of entropy increase is the difference between the entropy changes outside and inside: \[ \Delta S_{ ext{net}} = \Delta S_{ ext{outside}} - \Delta S_{ ext{inside}} = 77.31 - 67.75 \approx 9.56 \text{ J/K/s} \]

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Key Concepts

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

Heat Transfer
Heat transfer is a fundamental concept in thermodynamics that describes the movement of thermal energy from one place to another. In our context, the heat is transferring from the warmer inside of the house to the colder outside environment. This transfer happens because heat naturally flows from areas of higher temperature to lower temperature, striving for thermal equilibrium.
There are three main methods of heat transfer:
  • Conduction: Transfer of heat through direct contact of materials.
  • Convection: Transfer of heat through fluids (liquids or gases) where hot parts of the fluid rise and cooler ones sink.
  • Radiation: Transfer of heat via electromagnetic waves without involving particles.

In our exercise, we're primarily concerned with conduction and possibly convection as the house loses heat. Understanding the mechanisms of heat transfer helps in calculating the rate of entropy change, which quantifies how irreversibly energy dissipates.
Thermodynamics
Thermodynamics is the branch of physics that deals with heat and temperature, and their relation to energy and work. It encompasses several laws that describe how energy transfers and transforms.
The second law of thermodynamics is particularly relevant here. It states that the entropy of an isolated system will always increase over time. In simpler terms, systems naturally evolve towards thermodynamic equilibrium, where entropy, a measure of disorder, is maximized. This concept explains why heat moves from a warm area (inside the house) to a cooler one (outside) and helps in understanding entropy change calculations.
When calculating entropy, we apply these principles to determine how much disorder is introduced into the environment per second due to the heat leaking from the house.
Temperature Conversion
In the study of thermodynamics, precise temperature measurements are crucial. Often, temperatures are provided in Celsius but for scientific calculations involving heat and entropy, temperatures must be converted to Kelvin. The Kelvin scale starts at absolute zero, the theoretical lowest possible temperature where molecular motion ceases.
Converting Celsius to Kelvin is straightforward: just add 273.15 to the Celsius temperature.
  • For example, convert 22°C to Kelvin: \[ T_{\text{inside}} = 22 + 273.15 = 295.15 \text{ K} \]
  • Similarly, convert -14.5°C to Kelvin: \[ T_{\text{outside}} = -14.5 + 273.15 = 258.65 \text{ K} \]

Using Kelvin ensures that our entropy calculations are accurate, as entropy is a statistical measure that depends on extensive units like Kelvin. Always double-check your temperature conversions to avoid errors in related thermodynamics calculations.

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Most popular questions from this chapter

\- IP Engine A has an efficiency of \(66 \%\). Engine \(B\) absorbs the same amount of heat from the hot reservoir and exhausts twice as much heat to the cold reservoir. (a) Which engine has the greater efficiency? Explain. (b) What is the efficiency of engine B?

An air conditioner is used to keep the interior of a house at a temperature of \(21^{\circ} \mathrm{C}\) while the outside temperature is \(32^{\circ} \mathrm{C}\). If heat leaks into the house at the rate of \(11 \mathrm{kW},\) and the air conditioner has the efficiency of a Carnot engine, what is the mechanical power required to keep the house cool?

A freezer with a coefficient of performance of 3.88 is used to convert \(1.75 \mathrm{kg}\) of water to ice in one hour. The water starts at a temperature of \(20.0^{\circ} \mathrm{C}\), and the ice that is produced is cooled to a temperature of \(-5.00^{\circ} \mathrm{C}\). (a) How much heat must be removed from the water for this process to occur? (b) How much electrical energy does the freezer use during this hour of operation? \((\mathrm{c})\) How much heat is discarded into the room that houses the freezer?

A refrigerator with a coefficient of performance of 1.75 absorbs \(3.45 \times 10^{4} \mathrm{J}\) of heat from the low-temperature reservoir during each cycle. (a) How much mechanical work is required to operate the refrigerator for a cycle? (b) How much heat does the refrigerator discard to the high-temperature reservoir during each cycle?

To keep a room at a comfortable \(21.0^{\circ} \mathrm{C},\) a Carnot heat pump does 345 J of work and supplies it with 3240 J of heat. (a) How much heat is removed from the outside air by the heat pump? (b) What is the temperature of the outside air?

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