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A \(4 \mathrm{~L}\) jug of milk at \(25^{\circ} \mathrm{C}\) is placed in your refrigerator, where it is cooled down to \(5^{\circ} \mathrm{C}\). The high temperature in the Carnot refrigeration cycle is \(45^{\circ} \mathrm{C}\), the low temperature is \(-5^{\circ} \mathrm{C},\) and the properties of milk are the same as those of liquid water. Find the amount of energy that must be removed from the milk and the additional work needed to drive the refrigerator.

Short Answer

Expert verified
334.4 kJ of energy is removed and 62.34 kJ of work is needed.

Step by step solution

01

Convert temperatures to Kelvin

To calculate the energy and work in the Carnot cycle, we first need to convert all temperatures from degrees Celsius to Kelvin. The conversion formula is \( T(K) = T(°C) + 273.15 \). Thus:- The initial milk temperature is \( T_1 = 25 + 273.15 = 298.15 \, K \).- The cooled milk temperature is \( T_2 = 5 + 273.15 = 278.15 \, K \).- The high temperature of the cycle is \( T_H = 45 + 273.15 = 318.15 \, K \).- The low temperature of the cycle is \( T_L = -5 + 273.15 = 268.15 \, K \).
02

Calculate the energy removed from the milk

The energy removed from the milk is calculated using the formula for specific heat: \( Q = mc\Delta T \), where:- \( m = 4 \, kg \) (since density of water is \( 1 \, kg/L \) and jug capacity is \( 4 \, L \))- \( c = 4.18 \, kJ/kg\cdot K \) (specific heat capacity of water)- \( \Delta T = T_2 - T_1 = 278.15 - 298.15 = -20 \, K \)Thus, the energy removed is:\[ Q = 4 \, kg \times 4.18 \, kJ/kg\cdot K \times (-20 \, K) = -334.4 \, kJ \].The negative sign indicates energy removal.
03

Determine the efficiency of the Carnot refrigerator

The efficiency of a Carnot refrigerator (Coefficient of Performance, COP) is given by:\[ \text{COP} = \frac{T_L}{T_H - T_L} \]Where the temperatures are in Kelvin:\[ \text{COP} = \frac{268.15}{318.15 - 268.15} = \frac{268.15}{50} = 5.363 \].
04

Calculate the work required to remove the energy

The work done by the refrigerator can be determined using the COP formula:\[ \text{COP} = \frac{Q}{W} \quad \Rightarrow \quad W = \frac{Q}{\text{COP}} \]Substitute the values:\[ W = \frac{334.4 \, kJ}{5.363} \approx 62.34 \, kJ \].

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Key Concepts

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

Carnot Cycle
The Carnot cycle is a theoretical model that helps understand the efficiency of heat engines or refrigerators. It is named after the French engineer Sadi Carnot. The cycle represents an idealized process of heat exchange between two reservoirs at different temperatures. This cycle has four reversible stages:
  • Isothermal expansion
  • Adiabatic expansion
  • Isothermal compression
  • Adiabatic compression
Each stage within this cycle occurs under specific conditions. The isothermal processes maintain a constant temperature, whereas the adiabatic processes occur without any heat transfer to or from the system. Understanding the Carnot cycle is crucial because it provides the maximum possible efficiency that any heat engine or refrigerator can achieve between two temperatures by evaluating these processes.
Heat Transfer
Heat transfer is the process of thermal energy movement from one object or substance to another. In the context of the exercise, we are discussing how energy in the form of heat is transferred from the milk to the refrigerator. This energy transfer relies on the principle that heat naturally flows from a higher temperature to a lower temperature. There are three main modes of heat transfer:
  • Conduction: Transfer through physical contact.
  • Convection: Transfer through fluid movement.
  • Radiation: Transfer through electromagnetic waves.
In our scenario, energy is removed from milk (reducing its temperature) via a refrigerator by conduction, as the milk is placed directly inside the cooling chamber.
Coefficient of Performance
The Coefficient of Performance (COP) measures the efficiency of a refrigeration cycle. Different from efficiency in heat engines, COP refers to the ratio of the desired effect to the required input. For refrigerators, COP measures how much heat energy is removed compared to the amount of work done to remove it:\[\text{COP} = \frac{T_L}{T_H - T_L}\]Here, \( T_L \) is the low temperature inside the refrigerator, and \( T_H \) is the high temperature outside the refrigerator, both measured in Kelvin. The greater the COP, the more efficient the refrigeration cycle. In our calculation, the COP tells us that for every unit of work put into the refrigerator, 5.363 units of energy are removed from the milk, making it a highly efficient process.
Specific Heat Capacity
Specific heat capacity is a property of a substance that indicates how much energy is required to change its temperature by one degree Celsius (or one Kelvin). It is measured in units like kJ/kgâ‹…K. In this exercise, milk is treated as water with a specific heat capacity, \( c = 4.18 \) kJ/kgâ‹…K.The formula to calculate the change in thermal energy (or heat) when the temperature changes is:\[Q = mc\Delta T\]Where,
  • \( m \) is the mass of the substance (in kg),
  • \( c \) is the specific heat capacity,
  • \( \Delta T \) is the change in temperature.
This formula helps determine the heat energy removed from the milk as it cools from 25°C to 5°C. Such understanding is essential for engineering fields, as it applies in designing efficient heating or cooling systems.

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Most popular questions from this chapter

A car engine burns 5 kg fuel (equivalent to addition of \(Q_{H}\) ) at \(1500 \mathrm{~K}\) and rejects energy to the radiator and the exhaust at an average temperature of \(750 \mathrm{~K}\). If the fuel provides \(40000 \mathrm{~kJ} / \mathrm{kg}\), what is the maximum amount of work the engine can provide?

An air conditioner in a very hot region uses a power input of \(2.5 \mathrm{~kW}\) to \(\operatorname{cool}\) a \(5^{\circ} \mathrm{C}\) space with the high temperature in the cycle at \(40^{\circ} \mathrm{C}\). The \(Q_{H}\) is pushed to the ambient air at \(30^{\circ} \mathrm{C}\) in a heat exchanger where the transfer coefficient is \(50 \mathrm{~W} / \mathrm{m}^{2} \mathrm{~K}\). Find the required minimum heat transfer area.

If the efficiency of a power plant goes up as the low temperature drops, why do all power plants not reject energy at, say, \(-40^{\circ} \mathrm{C} ?\)

Air in a rigid \(1-\mathrm{m}^{3}\) box is at \(300 \mathrm{~K}, 200 \mathrm{kPa}\). It is heated to \(600 \mathrm{~K}\) by heat transfer from a reversible heat pump that receives energy from the ambient at \(300 \mathrm{~K}\) besides the work input. Use constant specific heat at \(300 \mathrm{~K}\). Since the COP changes, write \(d Q=m_{\text {air }} C_{\mathrm{v}} d T\) and find \(d W\). Integrate \(d W\) with the temperature to find the required heat pump work.

Helium has the lowest normal boiling point of any of the elements at \(4.2 \mathrm{~K}\). At this temperature the enthalpy of evaporation is \(83.3 \mathrm{~kJ} / \mathrm{kmol}\). A Carnot refrigeration cycle is analyzed for the production of \(1 \mathrm{kmol}\) of liquid helium at \(4.2 \mathrm{~K}\) from saturated vapor at the same temperature. What is the work input to the refrigerator and the COP for the cycle with an ambient at \(300 \mathrm{~K} ?\)

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