To calculate the work required to lower the temperature of an object, we need to consider how the coefficient of performance changes with the temperature of the object. (a) Find an expression for the work of cooling an object from \(\mathrm{T}\), to \(\mathrm{T}_{\mathrm{r}}\) when the refrigerator is in a room at a temperature \(\mathrm{T}_{\mathrm{h}}\). Hint. Write \(d w=d y l c\left(T\right.\) relate \(d q\) to \(\mathrm{dT}\) through the heat capacity \(\mathrm{C}_{p}\), and integrate the resulting expression. Assume that the heat capacity is independent of temperature in the range of interest. (b) Use the result in part (a) to calculate the work needed to freeze \(250 \mathrm{~g}\) of water in a refrigerator at \(293 \mathrm{~K}\). How long will it take when the refrigerator operates at \(100 \mathrm{~W}\) ?

Short Answer

Expert verified
The work required to freeze 250g of water from 293K to 273K in a refrigerator at 293K is calculated using the formula \(W = 250g \times (4.18 J/g\cdot K)(293K - 273K) \ln\left(\frac{293K}{273K}\right)\), and the time taken at 100W power can be found by \(t = \frac{W}{100W}\).

Step by step solution

01

Understand the Coefficient of Performance (COP)

The coefficient of performance (COP) for a refrigerator is defined as the ratio of the heat removed from the object (\(dQ\textsubscript{c}\)) to the work input (\(dW\textsubscript{in}\)). For an ideal refrigerator, the COP is given by \(COP = \frac{T_c}{T_h - T_c}\). Knowing this, we can relate the work to the heat removed.
02

Relate Heat to Temperature Change

Since the heat capacity (\(C_p\)) is constant, the heat removed from the object when its temperature decreases by a differential amount can be expressed by \(dQ = C_p dT\). The heat removed (\(dQ\textsubscript{c}\)) from the object is then given by the heat capacity times the infinitesimal change in temperature.
03

Express Work in Terms of Temperature Change

Now, we express the differential work (\(dW\textsubscript{in}\)) needed to remove the heat (\(dQ\textsubscript{c}\)) in terms of temperature. The work done by the refrigerator can be related to the heat extracted through the COP: \(dW = \frac{dQ}{COP} = \frac{C_p dT}{T_c/(T_h - T_c)} = \frac{C_p (T_h - T_c) dT}{T_c}\).
04

Integrate the Expression for Work

To find the total work for cooling the object from temperature \(T\) to \(T_r\), integrate the expression for \(dW\) between these two temperatures: \[W = \int_{T_r}^{T} \frac{C_p (T_h - T_c) dT}{T_c}\]. Since \(C_p\) is constant, it comes out of the integral, and \(T_c\), the temperature at which heat is removed, changes from \(T\) to \(T_r\). Hence the integral becomes: \[W = C_p (T_h - T) \int_{T_r}^{T} \frac{dT}{T}\] which simplifies to: \[W = C_p (T_h - T) \ln\left(\frac{T}{T_r}\right)\]
05

Apply the Expression to Freeze Water

Using the result from Step 4, calculate the work needed to freeze 250g of water from an initial temperature \(T\) to \(T_r\textsubscript{{ice}} = 273K\), within a refrigerator at \(T_h = 293K\). Using water's heat capacity \(C_p = 4.18 J/g\cdot K\) for \(T = 293K\): \[W = (4.18 J/g\cdot K)(293K - T) \ln\left(\frac{293K}{273K}\right)\]. Finally, estimate the total work for the 250g of water: \[W = 250g \times (4.18 J/g\cdot K)(293K - 273K) \ln\left(\frac{293K}{273K}\right)\]. Calculate this value for the total work.
06

Calculate the Time Required for Freezing

To find out how long it will take for the refrigerator to perform this work, we use the power rating of the refrigerator. If it operates at 100W (\(100 J/s\)), the time (\(t\)) is given by: \[ t = \frac{W}{Power} \]. Substitute the value of \(W\) from Step 5 and \(Power = 100W\) to find \(t\).

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 Vaia!

Key Concepts

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

Coefficient of Performance
In the realm of thermodynamics, particularly when dealing with refrigeration, the term Coefficient of Performance (COP) is crucial. It is a measure that represents the efficiency of a refrigeration system. The COP is defined as the ratio of the heat removed from a substance to the work input required to remove that heat. In mathematical terms, for an ideal refrigerator, the COP can be expressed as:
\[ COP = \frac{T_c}{T_h - T_c} \] where \(T_c\) is the temperature from which the heat is being removed (the cooling space) and \(T_h\) is the external temperature (often the room temperature). Understanding COP helps us appreciate how changes in temperature can affect the performance of a refrigeration system. Lower COP values indicate that more work is needed to remove a certain amount of heat, making the process less efficient.
Heat Capacity
When discussing the concept of heat capacity, we are referring to the amount of heat required to raise the temperature of a substance by one unit of temperature, often expressed in Joules per degree Celsius (J/°C) or Joules per Kelvin (J/K). Heat capacity is a fundamental property of matter that plays a pivotal role in thermodynamics and in calculating the energy changes associated with temperature variations. The relationship between heat capacity (\(C_p\)) and temperature change is linear, provided that \(C_p\) remains constant over the temperature range in question. This simplifies calculations as expressed in the equation:
\[ dQ = C_p dT \] For substances with a heat capacity that does not significantly change with temperature, energy calculations become more straightforward, allowing for easier integration to find the total heat exchange over a temperature interval.
Thermodynamic Temperature Change
The concept of thermodynamic temperature change is central to understanding heat transfer and the work involved in processes like cooling. Temperature change in a substance directly correlates to the exchange of heat energy - when a substance is cooled, its temperature decreases; conversely, heating a substance increases its temperature. The rate at which this temperature change occurs can affect the efficiency of cooling systems. In calculations, we often integrate the effect of a temperature change over the substance's entire mass to determine the total work or heat exchange. For instance, a larger temperature change typically requires more work to be done by the cooling system, stressing the importance of efficient performance for energy saving.
Energy Integration in Thermodynamics
The concept of energy integration in thermodynamics is a critical analytical tool that allows us to calculate the total amount of work or heat transfer over a process. In the context of cooling, integrating the work in terms of temperature change over the entire path from initial to final temperature provides the total work required for the process. In the case of our refrigeration example, the energy integration can be represented by the following integral:
\[ W = \int_{T_r}^{T} \frac{C_p (T_h - T_c) dT}{T_c} \] This integral accumulates the incremental amounts of work needed for each infinitesimal temperature change, giving us the complete work associated with cooling an object from temperature \(T\) to \(T_r\). By carrying out this integration, we can compute precise values for the energy required in thermodynamic processes, such as cooling water in a refrigerator.

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

Find an expression for the change in entropy when two blocks of the same substance and of equal mass, one at the temperature \(T_{\mathrm{h}}\) and the other at \(T_{c}\), are brought into thermal contact and allowed to reach equilibrium. Evaluate the change for two blocks of copper, each of mass \(500 \mathrm{~g}\), with \(C_{p, m}=24.4 \mathrm{~J} \mathrm{~K}^{-1}\) \(\mathrm{mol}^{-1}\), taking \(T_{\mathrm{h}}=500 \mathrm{~K}\) and \(T_{\mathrm{c}}=250 \mathrm{~K}\)

Show that, for a perfect gas, \((\partial U / \partial S)_{v}=T\) and \((\partial U / \partial V)_{\mathrm{S}}=-p\).

In biological cells, the energy released by the oxidation of foods (Impact on Biology \(12.2)\) is stored in adenosine triphosphate (ATP or \(\mathrm{ATP}^{4-}\) ). The essence of ATP's action is its ability to lose its terminal phosphate group by hydrolysis and to form adenosine diphosphate (ADP or ADP \(^{3-}\) ): $$ \mathrm{ATP}^{4-}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\mathrm{I}) \rightarrow \mathrm{ADP}^{3-}(\mathrm{aq})+\mathrm{HPO}_{4}^{2-}(\mathrm{aq})+\mathrm{H}_{3} \mathrm{O}^{+}(\mathrm{aq}) $$ At \(\mathrm{pH}=7.0\) and \(37^{\circ} \mathrm{C}\) (310 \(\mathrm{K}\), blood temperature) the enthalpy and Gibbs energy of hydrolysis are \(\Delta_{\mathrm{r}} H=-20 \mathrm{~kJ} \mathrm{~mol}^{-1}\) and \(\Delta_{\mathrm{r}} G=-31 \mathrm{~kJ} \mathrm{~mol}^{-1}\), respectively. Under these conditions, the hydrolysis of \(1 \mathrm{~mol} \Delta \mathrm{TP}^{4-}(\mathrm{aq})\) results in the extraction of up to \(31 \mathrm{~kJ}\) of energy that can be used to do nonexpansion work, such as the synthesis of proteins from amino acids, muscular contraction, and the activation of neuronal circuits in our brains. (a) Calculate and account for the sign of the entropy of hydrolysis of ATP at \(\mathrm{pH}=7.0\) and \(310 \mathrm{~K}\). (b) Suppose that the radius of a typical biological cell is \(10 \mu \mathrm{m}\) and that inside it \(10^{6}\) ATP molecules are hydrolysed each second. What is the power density of the cell in watts per cubic metre \(\left(1 \mathrm{~W}=1 \mathrm{~J} \mathrm{~S}^{-1}\right) ? \mathrm{~A}\) computer battery delivers about \(15 \mathrm{~W}\) and has a volume of \(100 \mathrm{~cm}^{3}\). Which has the greater power density, the cell or the battery? (c) The formation of glutamine from glutamate and ammonium ions requires \(14.2 \mathrm{~kJ} \mathrm{~mol}^{-1}\) of energy input. It is driven by the hydrolysis of ATP to ADP mediated by the enzyme glutamine synthetase. How many moles of ATP must be hydrolysed to form 1 mol glutamine?

The cycle involved in the operation of an internal combustion engine is called the Otto cycle. Air can be considered to be the working substance and can be assumed to be a perfect gas. The cycle consists of the following steps: (1) reversible adiabatic compression from \(\mathrm{A}\) to \(\mathrm{B},(2)\) reversible constantvolume pressure increase from \(\mathrm{B}\) to \(\mathrm{C}\) due to the combustion of a small amount of fuel, (3) reversible adiabatic expansion from \(\mathrm{C}\) to \(\mathrm{D}\), and \((4)\) reversible and constant-volume pressure decrease back to state A. Determine the change in entropy (of the system and of the surroundings) tor each step of the cycle and determine an expression for the efficiency of the cycle, assuming that the heat is supplied in Step \(2 .\) Evaluate the efficiency for a compression ratio of \(10: 1\). Assume that, in state \(\mathrm{A}, \mathrm{V}=4.00 \mathrm{dm}^{3}, p=1.00 \mathrm{~atm}\), and \(\mathrm{T}=300 \mathrm{~K}\), that \(\mathrm{V}_{\mathrm{A}}==10 \mathrm{~V}_{\mathrm{B}}, \mathrm{p}_{\mathrm{C}} / \mathrm{p}_{\mathrm{B}}=5\), and that \(C_{p m}=\frac{2}{2} R\).

The heat capacity of chloroform (trichloromethane, \(\mathrm{CHCl}_{3}\) ) in the range \(240 \mathrm{~K}\) to \(330 \mathrm{~K}\) is given by \(\mathrm{C}_{\mathrm{p}, \mathrm{m}} /\left(\mathrm{J} \mathrm{K}^{-1} \mathrm{~mol}^{-1}\right)=91.47+7.5 \times 10^{-2}(T / \mathrm{K})\). In a particular experiment, \(1.00 \mathrm{~mol} \mathrm{CHCl}_{3}\) is heated from \(273 \mathrm{~K}\) to \(300 \mathrm{~K}\). Calculate the change in molar entropy of the sample.

See all solutions

Recommended explanations on Chemistry 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.

Sign-up for free