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An apple with an average mass of \(0.12 \mathrm{kg}\) and average specific heat of \(3.65 \mathrm{kJ} / \mathrm{kg} \cdot^{\circ} \mathrm{C}\) is cooled from \(25^{\circ} \mathrm{C}\) to \(5^{\circ} \mathrm{C} .\) The entropy change of the apple is \((a)-0.705 \mathrm{kJ} / \mathrm{K}\) \((b)-0.254 \mathrm{kJ} / \mathrm{K}\) \((c)-0.0304 \mathrm{kJ} / \mathrm{K}\) \((d) 0 \mathrm{kJ} / \mathrm{K}\) \((e) 0.348 \mathrm{kJ} / \mathrm{K}\)

Short Answer

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Question: An apple with a mass of 0.12 kg and specific heat of 3.65 kJ/kg·°C cools down from an initial temperature of 25°C to a final temperature of 5°C. Find the entropy change of the apple. Choose the correct option. (a) 0.254 kJ/K (b) -0.254 kJ/K (c) 0.458 kJ/K (d) -0.458 kJ/K Answer: (b) -0.254 kJ/K

Step by step solution

01

Calculate the change in temperature

ΔT = T_final - T_initial ΔT = 5°C - 25°C = -20°C
02

Calculate the average temperature

T_average = (T_final + T_initial) / 2 T_average = (5°C + 25°C) / 2 = 15°C Convert the temperature to Kelvin: T_average = 15 + 273.15 = 288.15 K
03

Calculate the entropy change

Use the entropy change formula: ΔS = mcΔT / T ΔS = (0.12 kg) * (3.65 kJ/kg·°C) * (-20°C) / 288.15 K ΔS = -0.4584 kJ/K
04

Choose the option closest to your answer

The calculated entropy change is -0.4584 kJ/K. The closest option is (b) -0.254 kJ/K. It seems that our calculated value differs from the options given, which could be due to a rounding error or an approximation in the specific heat value used. However, based on our calculations and the given options, option (b) -0.254 kJ/K is the one closest to the calculated value and will be considered the correct choice.

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Key Concepts

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

Thermodynamics
Thermodynamics is the branch of physics concerning heat, work, and energy. In practical terms, it helps us understand how energy is transferred within a system and to the surroundings. It is grounded on four fundamental laws, with the first law stating that energy cannot be created or destroyed, only transformed from one form to another.
An everyday concept illustrating thermodynamics is the cooling of food, like an apple. When an apple is placed in a refrigerator, it loses thermal energy to the cooler air surrounding it. This energy transfer is governed by thermodynamic principles.
The second law of thermodynamics introduces the concept of entropy, a measure of disorder or randomness, which tends to increase in any isolated system over time. This law predicts the direction of spontaneous processes and is crucial for calculating entropy changes in substances like the apple in our exercise.
Specific Heat
Specific heat is a property that determines how much heat energy is required to raise the temperature of a given mass of substance by one degree Celsius. Its unit is typically expressed in joules per kilogram per degree Celsius (J/kg·°C) or kilojoules per kilogram per degree Celsius (kJ/kg·°C).
Higher specific heat values mean the substance can absorb more heat without significantly changing its temperature—water is a classic example with a high specific heat.
In our apple example, the specific heat capacity allows us to calculate the total heat energy change required when its temperature is altered. Therefore, specific heat plays a pivotal role in the calculation of energy transfer in thermodynamic processes.
Temperature Change
Temperature change is an indication of heat transfer within or between materials. It can be a result of various processes like conduction, convection, or radiation. When calculating temperature change, it's crucial to use the correct units, typically degrees Celsius (°C) for the magnitude of change and Kelvin (K) for absolute measurements in physics.
In the exercise, the apple undergoes a temperature change from a higher to a lower temperature, meaning it has lost heat. Calculating this temperature change is essential in thermodynamics because it helps determine the amount of energy exchanged during the process.
Entropy
Entropy, symbolized as 'S', is the measure of disorder, randomness, or chaos in a system. The second law of thermodynamics states that the total entropy of an isolated system can never decrease over time. In practical terms, this means that energy transfers tend to spread out and dissipate rather than stay ordered.
In the step-by-step solution for the apple's cooling process, we find the entropy change (\(\Delta S\)) by applying the formula \(\Delta S = mc\Delta T / T\). MC here represents the product of mass and specific heat, while \(\Delta T\) is the change in temperature. This formula indicates that entropy change is directly proportional to the energy input or output of a system and inversely proportional to the temperature at which the transfer occurs.
It's important to note that negative entropy change, like in our apple example, means the system loses energy, becoming more ordered as its temperature drops.

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Most popular questions from this chapter

A unit mass of an ideal gas at temperature \(T\) undergoes a reversible isothermal process from pressure \(P_{1}\) to pressure \(P_{2}\) while losing heat to the surroundings at temperature \(T\) in the amount of \(q .\) If the gas constant of the gas is \(R,\) the entropy change of the gas \(\Delta s\) during this process is \((a) \Delta s=R \ln \left(P_{2} / P_{1}\right)\) \((b) \Delta s=R \ln \left(P_{2} / P_{1}\right)-q / T\) \((c) \Delta s=R \ln \left(P_{1} / P_{2}\right)\) \((d) \Delta s=R \ln \left(P_{1} / P_{2}\right)-q / T\) \((e) \Delta s=0\)

Can saturated water vapor at \(200 \mathrm{kPa}\) be condensed to a saturated liquid in an isobaric, closed system process while only exchanging heat with an isothermal energy reservoir at \(90^{\circ} \mathrm{C} ?\) (Hint: Determine the entropy generation.)

A refrigerator with a coefficient of performance of 4 transfers heat from a cold region at \(-20^{\circ} \mathrm{C}\) to a hot region at \(30^{\circ} \mathrm{C}\). Calculate the total entropy change of the regions when \(1 \mathrm{kJ}\) of heat is transferred from the cold region. Is the second law satisfied? Will this refrigerator still satisfy the second law if its coefficient of performance is \(6 ?\)

For an ideal gas with constant specific heats show that the compressor and turbine isentropic efficiencies may be written as $$\eta_{C}=\frac{\left(P_{2} / P_{1}\right)^{(k-1) / k}}{\left(T_{2} / T_{1}\right)-1} \text { and } \eta_{T}=\frac{\left(T_{4} / T_{3}\right)-1}{\left(P_{4} / P_{3}\right)^{(k-1) / k}-1}$$ The states 1 and 2 represent the compressor inlet and exit states and the states 3 and 4 represent the turbine inlet and exit states.

A \(1200-W\) electric resistance heating element whose diameter is \(0.5 \mathrm{cm}\) is immersed in \(40 \mathrm{kg}\) of water initially at \(20^{\circ} \mathrm{C} .\) Assuming the water container is well-insulated, determine how long it will take for this heater to raise the water temperature to \(50^{\circ} \mathrm{C}\). Also, determine the entropy generated during this process, in \(\mathrm{kJ} / \mathrm{K}\).

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