/*! 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 55 Assume that it takes \(0.0700 \m... [FREE SOLUTION] | 91影视

91影视

Assume that it takes \(0.0700 \mathrm{~J}\) of energy to heat a \(1.00-\mathrm{g}\) sample of mercury from \(10.000^{\circ} \mathrm{C}\) to \(10.500{ }^{\circ} \mathrm{C}\) and that the heat capacity of mercury is constant, with a negligible change in volume as a function of temperature. Find the change in entropy if this sample is heated from \(10 .{ }^{\circ} \mathrm{C}\) to \(100^{\circ} \mathrm{C}\).

Short Answer

Expert verified
Question: Calculate the change in entropy of a 1.00-g sample of mercury when it is heated from 10.0掳C to 100.0掳C, given that it takes 0.0700 J of energy to heat the sample from 10.0掳C to 10.5掳C and assuming a constant heat capacity. Answer: The change in entropy of the mercury sample when heated from 10.0掳C to 100.0掳C is approximately 0.175 J K鈦宦.

Step by step solution

01

Determine the heat capacity of mercury

Given that it takes 0.0700 J of energy to heat a 1.00-g sample of mercury from 10.0掳C to 10.5掳C, we can determine the specific heat capacity (c) of mercury using the formula: q = mc螖T Where: q = heat energy (0.0700 J), m = mass of the sample (1.00 g), c = specific heat capacity, and 螖T = change in temperature (0.5 K). Rearranging the formula for c, we get: c = q/(m螖T) Substituting the given values, we find the specific heat capacity of mercury: c = 0.0700 J / (1.00 g * 0.5 K) = 0.140 J g鈦宦 K鈦宦
02

Calculate the change in entropy

To find the change in entropy (螖S) for heating the mercury sample from 10.0掳C to 100.0掳C, we use the following formula: 螖S = mc\int_{T_1}^{T_2} \frac{dT}{T} Where: 螖S = change in entropy, m = mass of the sample (1.00 g), c = specific heat capacity (0.140 J g鈦宦 K鈦宦), T_1 = initial temperature (10.0掳C = 283.15 K), and T_2 = final temperature (100.0掳C = 373.15 K), In this case, c is constant, so we can take it out of the integral: 螖S = mc (ln(T_2) - ln(T_1)) Substituting the values we found, we can now calculate the change in entropy: 螖S = (1.00 g)(0.140 J g鈦宦 K鈦宦) (ln(373.15 K) - ln(283.15 K)) 螖S 鈮 0.140 J K鈦宦 (1.2507) = 0.175 J K鈦宦 The change in entropy of the mercury sample when heated from 10.0掳C to 100.0掳C is approximately 0.175 J K鈦宦.

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 91影视!

Key Concepts

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

Specific Heat Capacity
Understanding the concept of specific heat capacity is crucial for solving problems involving temperature changes and energy transfer. Specific heat capacity, often just called specific heat, is defined as the amount of heat required to raise the temperature of one gram of a substance by one degree Celsius (or one kelvin).

It is an intrinsic property of a substance, meaning it doesn't change with the amount of substance you have; whether you have 1 gram or 1000 grams, the specific heat capacity will be the same for a given substance under consistent conditions. To grasp this idea, imagine heating up different materials, like water and metal. You might have noticed that metal heats up faster compared to water. This is because metals typically have a lower specific heat capacity than water, requiring less energy to raise their temperature.

In thermodynamics, the specific heat capacity is a gateway to understanding energy transfer. When applied to the dimension of entropy change calculation, specific heat tells us how much energy is dispersed or drawn in during a temperature change, which we interpret as a measure of disorder or randomness in the system.
Thermodynamics
Thermodynamics is a branch of physics that studies the movement of heat and its conversion to different forms of energy. At its heart are the laws of thermodynamics which govern these energy transformations. One fundamental concept here is entropy, often described as the measure of the disorder or randomness in a system. The second law of thermodynamics states that for any isolated system, entropy can only increase, explaining why processes are irreversible and why we can't completely convert all heat energy to useful work.

In the context of our exercise, thermodynamics helps us understand how heat energy relates to changes in entropy. When heat is added to a system, like our mercury sample, it tends to disperse the energy throughout the system, increasing its overall entropy. Calculating this change provides insight into the 'spread' of energy and helps decode the complexity of thermal processes.
Integral Calculus in Physics
In the world of physics, integral calculus is a powerful tool used to calculate quantities that involve summation over continuous variables. Integrals, for instance, can tell us the total energy work done along a path or the change in a physical quantity over time or space. When it comes to thermodynamics, integrals become crucial in calculating the total entropy change over a range of temperatures as we've seen in the textbook exercise.

To find the change in entropy, we performed an integral over the temperature range with the formula involving the natural logarithm function. This approach is made possible by integral calculus and allows us to summarize the continuous increase in entropy at each infinitesimal step as the temperature rises from the initial to the final state. It's the integral's calculation that yields the total change in entropy in a neat, quantifiable number, reflecting the cumulative effect of the heat transfer.

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 heat engine consists of a heat source that causes a monatomic gas to expand, pushing against a piston, thereby doing work. The gas begins at a pressure of \(300 . \mathrm{kPa}\), a volume of \(150 . \mathrm{cm}^{3}\), and room temperature, \(20.0^{\circ} \mathrm{C}\). On reaching a volume of \(450 . \mathrm{cm}^{3}\), the piston is locked in place, and the heat source is removed. At this point, the gas cools back to room temperature. Finally, the piston is unlocked and used to isothermally compress the gas back to its initial state. a) Sketch the cycle on a \(p V\) -diagram. b) Determine the work done on the gas and the heat flow out of the gas in each part of the cycle. c) Using the results of part (b), determine the efficiency of the engine.

20.14 Imagine dividing a box into two equal parts, part \(A\) on the left and part \(B\) on the right. Four identical gas atoms, numbered 1 through 4 , are placed in the box. What are most probable and second most probable distributions (for example, 3 atoms in \(\mathrm{A}, 1\) atom in \(\mathrm{B}\) ) of gas atoms in the box? Calculate the entropy, \(S\), for these two distributions. Note that the configuration with 3 atoms in \(\mathrm{A}\) and 1 atom in \(\mathrm{B}\) and that with 1 atom in A and three atoms in B count as different configurations.

An outboard motor for a boat is cooled by lake water at \(15.0^{\circ} \mathrm{C}\) and has a compression ratio of \(10.0 .\) Assume that the air is a diatomic gas. a) Calculate the efficiency of the engine's Otto cycle. b) Using your answer to part (a) and the fact that the efficiency of the Carnot cycle is greater than that of the Otto cycle, estimate the maximum temperature of the engine.

What capacity must a heat pump with a coefficient of performance of 3 have to heat a home that loses heat energy at a rate of \(12 \mathrm{~kW}\) on the coldest day of the year? a) \(3 \mathrm{~kW}\) c) \(10 \mathrm{~kW}\) e) \(40 \mathrm{~kW}\) b) \(4 \mathrm{~kW}\) d) \(30 \mathrm{~kW}\)

Consider a room air conditioner using a Carnot cycle at maximum theoretical efficiency and operating between the temperatures of \(18^{\circ} \mathrm{C}\) (indoors) and \(35^{\circ} \mathrm{C}\) (outdoors). For each 1.00 J of heat flowing out of the room into the air conditioner: a) How much heat flows out of the air conditioner to the outdoors? b) By approximately how much does the entropy of the room decrease? c) By approximately how much does the entropy of the outdoor air increase?

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.