/*! 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 58 A restaurant refrigerator has a ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A restaurant refrigerator has a coefficient of performance of 4.6. If the temperature in the kitchen outside the refrigerator is 32\(^\circ\)C, what is the lowest temperature that could be obtained inside the refrigerator if it were ideal?

Short Answer

Expert verified
The lowest temperature inside the ideal refrigerator is approximately -218.66 °C.

Step by step solution

01

Understand the Coefficient of Performance

The coefficient of performance (COP) for a refrigerator is defined as the ratio of heat removed from the cold reservoir, divided by the work input to the refrigerator. The formula for COP is: \( \text{COP} = \frac{Q_c}{W} = \frac{T_c}{T_h - T_c} \), where \(Q_c\) is the heat removed from the cold reservoir, \(W\) is the work done, \(T_c\) is the temperature of the cold reservoir (inside the refrigerator), and \(T_h\) is the temperature of the hot reservoir (the kitchen temperature), all in Kelvin.
02

Convert Temperatures to Kelvin

To work with temperatures in this formula, we need to convert the given kitchen temperature to Kelvin. \( T_h = 32 + 273.15 = 305.15 \text{ K} \).
03

Apply the COP Formula

Rearranging the COP formula to solve for the cold temperature, we have: \( T_c = \frac{T_h}{1 + \text{COP}} \). Substitute the known values: \( T_c = \frac{305.15}{1 + 4.6} \).
04

Perform the Calculations

Calculate \( T_c \) using the rearranged formula: \( T_c = \frac{305.15}{5.6} \approx 54.49 \text{ K} \).
05

Convert the Result to Celsius

Convert the calculated Kelvin temperature back to Celsius by subtracting 273.15: \( 54.49 - 273.15 = -218.66 \text{ °C} \).

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.

Refrigeration
Refrigeration is a process of removing heat from a designated area and maintaining a substantially lower temperature compared to its surroundings. In the context of this exercise, the refrigerator is located in a kitchen, extracting heat from inside the fridge to keep its contents cold. The efficiency of this heat removal process is gauged by the coefficient of performance (COP).
The refrigerator does not cool by making things 'cold'; instead, it removes heat. The COP in refrigeration acts as a measure of effectiveness. A higher COP indicates that less work is required to remove a certain amount of heat, which is energy-efficient. In our problem, the refrigerator has a COP of 4.6, implying that it's quite efficient in terms of energy usage, as it performs 4.6 units of cooling for every unit of work it consumes. Understanding this makes it easier to determine how effective a refrigeration process will be.
Kelvin Temperature Conversion
In thermodynamics and especially in refrigeration problems, temperature is often expressed in Kelvin (K) rather than Celsius (°C) or Fahrenheit because Kelvin is an absolute temperature scale where zero is absolute zero, the point where no thermal energy remains in a substance.
To convert temperatures from Celsius to Kelvin, which is necessary for calculations involving thermodynamic formulas such as the COP, you simply add 273.15 to the Celsius temperature. For example, the kitchen temperature given in the problem is 32°C. By adding 273.15, you convert it to 305.15 K.
This conversion is crucial when using formulas like the COP, since both input and output temperatures in these formulas should be in Kelvin to ensure accuracy in calculations. Any mistake in this conversion can lead to incorrect results.
Thermodynamics
Thermodynamics is the branch of physics concerned with heat and temperature and their relation to energy and work. It encompasses concepts like energy transfer, the laws of thermodynamics, and the efficiency of thermal machines such as engines and refrigerators.
In thermodynamics, the first law essentially states that energy cannot be created or destroyed—only transformed from one form to another. For our refrigerator, this means that the work input into the refrigerator's motor is transformed into heat, which is then moved or removed from the inside of the fridge to the outside environment (the kitchen).
The relationship between the heat removed from the cold region, the work input, and the temperatures of the different spaces is what defines the COP. Understanding thermodynamic principles helps explain why refrigerators can cool spaces efficiently, like in the exercise where the process allows the refrigerator to reach very low temperatures inside despite the warmer kitchen environment.

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 car engine whose output power is 135 hp operates at about 15\(\%\) efficiency. Assume the engine's water temperature of 85\(^\circ\)C is its cold-temperature (exhaust) reservoir and 495\(^\circ\)C is its thermal "intake" temperature (the temperature of the exploding gas\(-\)air mixture). (\(a\)) What is the ratio of its efficiency relative to its maximum possible (Carnot) efficiency? (\(b\)) Estimate how much power (in watts) goes into moving the car, and how much heat, in joules and in kcal, is exhausted to the air in 1.0 h.

(II) A Carnot engine's operating temperatures are 210\(^\circ\)C and 45\(^\circ\)C. The engine's power output is 910W. Calculate the rate of heat output.

(II) A 1.0-L volume of air initially at 3.5 atm of (gauge) pressure is allowed to expand isothermally until the pressure is 1.0 atm. It is then compressed at constant pressure to its initial volume, and lastly is brought back to its original pressure by heating at constant volume. Draw the process on a \(PV\) diagram, including numbers and labels for the axes.

When 5.80 \(\times 10^5\) J of heat is added to a gas enclosed in a cylinder fitted with a light frictionless piston maintained at atmospheric pressure, the volume is observed to increase from 1.9 m\(^3\) to 4.1 m\(^3\). Calculate (\(a\)) the work done by the gas, and (\(b\)) the change in internal energy of the gas. (\(c\)) Graph this process on a \(PV\) diagram.

A cooling unit for a new freezer has an inner surface area of 8.0 m\(^2\) and is bounded by walls 12 cm thick with a thermal conductivity of 0.050 W/m\(\cdot\)K. The inside must be kept at \(-15^\circ\)C in a room that is at 22\(^\circ\)C. The motor for the cooling unit must run no more than 15\(\%\) of the time. What is the minimum power requirement of the cooling motor?

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.