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Calculate the change in entropy of \(250 \mathrm{g}\) of water heated slowly from \(20.0^{\circ} \mathrm{C}\) to \(80.0^{\circ} \mathrm{C} .\) (Suggestion: Note that \(d Q=m c d T)\)

Short Answer

Expert verified
After calculation, the change in entropy of the water is found to be approximately 111.05 J/K.

Step by step solution

01

Identify Given Variables

Firstly, list down the given values. Here, the mass (m) of the water is 250g or 0.25kg. The initial temperature \(T_1\) is 20°C or 293.15K (since Kelvin scale is used in these calculations). The final temperature \(T_2\) is 80°C or 353.15K. Also, the specific heat capacity of water (c) is a universally accepted value of 4186 J/kg∙K.
02

Apply Entropy Change Formula

Next, utilize the formula for calculating change in entropy, which is \( \Delta S = \int_{T_1}^{T_2} \frac{dQ}{T} = \int_{T_1}^{T_2} \frac{mc \, dT}{T} \).
03

Calculate the Integral

Now, calculate the definite integral \( \Delta S = m c \int_{T_1}^{T_2} \frac{d T}{T} \). This is a rather straightforward calculation if you recall the integral of \(1/x\) is \(ln|x|\), and so this solves as \( \Delta S = m c [ln(T_2) - ln(T_1)] = m c ln(\frac{T_2}{T_1}) \).
04

Substituting the Values

Finally, substitute the given values into the calculated formula to compute the entropy. \( \Delta S = 0.25 kg × 4186 J/ kg∙K × ln(353.15/ 293.15) \)

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Key Concepts

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

Thermodynamics
Thermodynamics is a branch of physics that deals with the relationships and conversions between heat and other forms of energy. Entropy, a core concept in thermodynamics, measures the amount of disorder or randomness in a system. During a process where heat is transferred, the entropy of a system tends to increase. This is especially evident when we heat a substance, like water, causing the water molecules to move more vigorously and become more disordered.

In the context of the given exercise, the increase in entropy reflects the greater disorder as water is heated from a lower temperature to a higher temperature. Notably, the second law of thermodynamics states that the total entropy of an isolated system can never decrease over time. In practical terms for the exercise, when we apply heat to water, we expect an increase in entropy because the water molecules are absorbing energy and spreading it within, increasing their disorder.
Specific Heat Capacity
The specific heat capacity, denoted as 'c' in physics equations, is the amount of heat energy required to raise the temperature of one kilogram of a substance by one degree Celsius or Kelvin. Each substance has a unique specific heat capacity, which is a measure of how much energy the substance can store. Water, with a high specific heat capacity of 4186 J/kg∙K, can absorb a lot of heat before its temperature rises significantly.

This property of water is crucial in the problem we are solving as it dictates how much the entropy changes for a given mass of water when heated. Due to the high heat capacity, a significant amount of heat can be transferred into the water with a relatively modest increase in temperature, leading to an increase in entropy, as we've calculated in the exercise. A substance with a lower specific heat capacity would experience a smaller change in entropy when the same amount of heat is applied.
Integral Calculus in Physics
Integral calculus is a part of mathematics that deals with the accumulation of quantities and how they change continuously. It plays a pivotal role in physics, particularly in thermodynamics, where it helps in calculating quantities like work, heat, and, as in our example, entropy. The integral sign (∫) represents the summing up of infinitesimally small changes over a range.

In the exercise, we use integral calculus to calculate the entropy change by integrating the amount of heat added over the temperature range through which the water is heated. This is represented mathematically as \( \Delta S = m c \int_{T_1}^{T_2} \frac{d T}{T} \), which produces the logarithmic expression due to the integral of 1/T. This integral is essential for finding the total change in entropy as the water moves from one steady state to another and is reflective of the continuous nature of the heat transfer in the process.

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Most popular questions from this chapter

A heat engine operates between two reservoirs at \(T_{2}=600 \mathrm{K}\) and \(T_{1}=350 \mathrm{K} .\) It takes in \(1000 \mathrm{J}\) of energy from the higher-temperature reservoir and performs 250 J of work. Find (a) the entropy change of the Universe \(\Delta S_{U}\) for this process and (b) the work \(W\) that could have been done by an ideal Carnot engine operating between these two reservoirs. (c) Show that the difference between the amounts of work done in parts (a) and (b) is \(T_{1} \Delta S_{U}\)

One of the most efficient heat engines ever built is a steam turbine in the Ohio valley, operating between \(430^{\circ} \mathrm{C}\) and \(1870^{\circ} \mathrm{C}\) on energy from West Virginia coal to produce electricity for the Midwest. (a) What is its maximum theoretical efficiency? (b) The actual efficiency of the engine is \(42.0 \% .\) How much useful power does the engine deliver if it takes in \(1.40 \times 10^{5} \mathrm{J}\) of energy each second from its hot reservoir?

How much work is required, using an ideal Carnot refrigerator, to change \(0.500 \mathrm{kg}\) of tap water at \(10.0^{\circ} \mathrm{C}\) into ice at \(-20.0^{\circ} \mathrm{C} ?\) Assume the temperature of the freezer compartment is held at \(-20.0^{\circ} \mathrm{C}\) and the refrigerator exhausts energy into a room at \(20.0^{\circ} \mathrm{C}\)

A refrigerator maintains a temperature of \(0^{\circ} \mathrm{C}\) in the cold compartment with a room temperature of \(25.0^{\circ} \mathrm{C}\). It removes energy from the cold compartment at the rate of \(8000 \mathrm{kJ} / \mathrm{h} .\) (a) What minimum power is required to operate the refrigerator? (b) The refrigerator exhausts energy into the room at what rate?

What is the maximum possible coefficient of performance of a heat pump that brings energy from outdoors at \(-3.00^{\circ} \mathrm{C}\) into a \(22.0^{\circ} \mathrm{C}\) house? Note that the work done to run the heat pump is also available to warm up the house.

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