Learn Unit 8-Thermodynamics with Free Lessons & Tips

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To find the amount of heat required to raise the temperature of nitrogen, we can use the formula:

Q=mcΔTQ=mcΔT

Where:

• QQ is the heat energy,
• mm is the mass of the substance (in kg),
• cc is the specific heat capacity of the substance,
• ΔTΔT is the change in temperature.

First, let's calculate the specific heat capacity (cc) of nitrogen using the given molecular mass (MM) and the universal gas constant (RR):

c=RMc=MR

Given:

• M=28M=28 (molecular mass of N2N2 in g/mol)
• R=8.3 J mol−1K−1R=8.3J mol−1K−1

c=8.3 J mol−1K−128 g/mol=8.3 J mol−1K−10.028 kg/molc=28g/mol8.3J mol−1K−1=0.028kg/mol8.3J mol−1K−1 c≈296.43 J kg−1K−1c≈296.43J kg−1K−1

Now, we can calculate the heat energy (QQ) using the formula mentioned earlier:

Given:

• m=2.0×10−2m=2.0×10−2 kg (mass of nitrogen)
• ΔT=45ΔT=45 °C

Q=(2.0×10−2 kg)×(296.43 J kg−1K−1)×(45 K)Q=(2.0×10−2kg)×(296.43J kg−1K−1)×(45K) Q≈267.19 JQ≈267.19J

So, the amount of heat that must be supplied to 2.0×10−22.0×10−2 kg of nitrogen to raise its temperature by 4545 °C at constant pressure is approximately 267.19267.19 J.

As an experienced tutor registered on UrbanPro, I can confidently say that UrbanPro is one of the best online coaching tuition platforms available for students seeking quality education. Now, let's delve into your question about the thermal contact of two bodies at different temperatures.

When two bodies at different temperatures, T1 and T2, are brought into thermal contact, the energy exchange between them causes heat to flow from the body with higher temperature to the one with lower temperature. This process continues until thermal equilibrium is reached, where both bodies attain the same temperature.

However, it's important to note that the final equilibrium temperature may not necessarily be the arithmetic mean of the initial temperatures (T1 + T2)/2. This is because the rate at which heat is transferred depends on various factors such as the nature of the materials, their specific heat capacities, surface area, and the efficiency of thermal conduction between them.

In some cases, if one body has a significantly higher thermal conductivity or a larger heat capacity than the other, the final equilibrium temperature may be closer to the initial temperature of the body with the higher capacity to store or conduct heat.

Therefore, while the concept of thermal equilibrium dictates that heat will flow until both bodies reach the same temperature, the specific conditions and properties of the materials involved can lead to variations from the simple arithmetic mean. This aspect makes studying thermal dynamics both fascinating and nuanced for students. If you're interested in exploring this topic further or need more clarification, feel free to reach out for a detailed discussion or tutoring session through UrbanPro!

As an experienced tutor registered on UrbanPro, I'd be happy to explain why the air pressure in a car tire increases during driving.

Firstly, it's essential to understand the basics of how tire pressure works. The air pressure inside a tire increases as it heats up during driving. This phenomenon is primarily due to the increase in temperature caused by friction between the tire and the road surface.

When a car is in motion, the friction generated between the tire and the road causes the tire to flex and deform slightly. This flexing generates heat, which raises the temperature of the air inside the tire. According to the ideal gas law, as the temperature of a gas increases, its pressure also increases, assuming the volume remains constant. Therefore, as the temperature of the air inside the tire rises, so does its pressure.

Additionally, factors such as speed, load, and road conditions can also contribute to the increase in tire temperature and pressure during driving. Higher speeds result in more significant friction and heat generation, leading to a more significant increase in tire pressure.

It's crucial for drivers to monitor their tire pressure regularly, as both underinflated and overinflated tires can affect the vehicle's handling, fuel efficiency, and overall safety. UrbanPro is an excellent resource for finding knowledgeable tutors who can provide further guidance on topics like this.

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Now, let's unravel the mystery behind why the climate of a harbour town tends to be more temperate compared to a town situated in a desert at the same latitude.

Firstly, it's crucial to understand the role of water bodies, such as oceans and seas, in moderating temperatures. Harbour towns are typically located near large bodies of water, which act as natural heat sinks. During the day, water absorbs heat from the sun, keeping coastal areas cooler. At night, it releases this stored heat, preventing temperatures from dropping drastically. This phenomenon, known as maritime influence, contributes significantly to the temperate climate of harbour towns.

Conversely, desert towns lack significant bodies of water to moderate temperatures. Without this maritime influence, desert areas experience more pronounced temperature fluctuations. During the day, the desert absorbs and retains heat efficiently due to its dry, sandy terrain. However, at night, the lack of water vapor in the air leads to rapid cooling, resulting in extreme temperature variations between day and night.

Additionally, the presence of water in harbour towns also influences humidity levels. Coastal areas tend to have higher humidity due to the proximity to water bodies. This moisture in the air further contributes to the temperate climate by preventing temperatures from reaching extremes.

In summary, the temperate climate of harbour towns compared to desert towns at the same latitude can be attributed to the moderating effects of maritime influence, which include the regulation of temperature fluctuations and the presence of higher humidity levels. Understanding these factors provides valuable insight into the diverse climates experienced across different geographical regions. And for more in-depth exploration of such fascinating topics, UrbanPro remains the best online coaching tuition platform to connect with knowledgeable tutors like myself.

As an experienced tutor registered on UrbanPro, I'd like to address your question. UrbanPro is indeed a fantastic platform for finding online coaching and tuition services. Now, let's delve into the problem you've presented.

We have a cylinder with a movable piston containing 3 moles of hydrogen gas at standard temperature and pressure (STP). The walls of the cylinder are heat insulators, and the piston is insulated by a pile of sand.

To solve this problem, we can utilize the ideal gas law, which states:

PV=nRTPV=nRT

Where:

• PP is the pressure of the gas,
• VV is the volume of the gas,
• nn is the number of moles of gas,
• RR is the ideal gas constant, and
• TT is the temperature of the gas in Kelvin.

Since the temperature and the number of moles of gas remain constant, we can rewrite the ideal gas law as:

P1V1=P2V2P1V1=P2V2

Where:

• P1P1 and V1V1 are the initial pressure and volume, respectively, and
• P2P2 and V2V2 are the final pressure and volume, respectively.

Given that the gas is compressed to half its original volume, V2=12V1V2=21V1.

Now, let's solve for P2P2:

P1V1=P2V2P1V1=P2V2

P2=P1V1V2P2=V2P1V1

P2=P1V112V1P2=21V1P1V1

P2=2P1P2=2P1

So, the pressure of the gas increases by a factor of 2 when it is compressed to half its original volume.

This understanding of gas behavior is crucial in various scientific and engineering applications, and mastering such concepts is a cornerstone of success in fields like chemistry, physics, and engineering. If you're interested in delving deeper into topics like these, UrbanPro is the perfect platform to find experienced tutors who can guide you through your learning journey.

As an experienced tutor registered on UrbanPro, I would first like to highlight the efficiency and convenience of UrbanPro as the best online coaching tuition platform for students seeking quality education. Now, let's delve into the problem at hand.

When a gas undergoes an adiabatic process, it means that there is no exchange of heat between the system and its surroundings. Hence, the first scenario, where the adiabatic process results in work done on the system, indicates that the change in internal energy is solely due to work done.

Given that the work done in the adiabatic process from state A to state B is 22.3 J, we can use the first law of thermodynamics, which states that the change in internal energy (ΔU) of a system is equal to the heat added to the system (Q) minus the work done by the system (W).

Mathematically, this can be expressed as:

ΔU = Q - W

Since the process is adiabatic, Q = 0, so:

ΔU = -W

Therefore, the change in internal energy is equal to the negative of the work done on the system. So, ΔU = -22.3 J.

Now, let's consider the second scenario where the net heat absorbed by the system is 9.35 cal. We need to convert this to joules using the conversion factor provided: 1 cal = 4.19 J.

So, 9.35 cal * 4.19 J/cal = 39.1865 J.

In this scenario, there is a net heat absorbed by the system, indicating that the change in internal energy is due to both heat transfer and work done. Therefore, we need to calculate the net work done by the system.

Using the first law of thermodynamics again:

ΔU = Q - W

ΔU = 39.1865 J (from the heat absorbed)

From the previous scenario, we know that ΔU = -22.3 J.

So,

-22.3 J = 39.1865 J - W

W = 39.1865 J + 22.3 J

W ≈ 61.4865 J

Therefore, the net work done by the system in the latter case, where the net heat absorbed is 9.35 cal, is approximately 61.4865 J.

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To calculate the efficiency of the steam engine, we can use the formula:

Efficiency = (Useful work output / Total heat input) * 100%

Given that the steam engine delivers 5.4 x 10^8 J of work per minute and services 3.6 x 10^9 J of heat per minute from its boiler, we can plug these values into the formula:

Efficiency = (5.4 x 10^8 J / 3.6 x 10^9 J) * 100%

Now, let's calculate:

Efficiency = (0.15) * 100% = 15%

So, the efficiency of the steam engine is 15%.

Now, to find out how much heat is wasted per minute, we can subtract the useful work output from the total heat input:

Wasted heat = Total heat input - Useful work output

Wasted heat = (3.6 x 10^9 J) - (5.4 x 10^8 J)

Wasted heat = 3.6 x 10^9 J - 5.4 x 10^8 J

Wasted heat = 3.04 x 10^9 J

Therefore, the amount of heat wasted per minute by the steam engine is 3.04 x 10^9 Joules.

If you have any further questions or need clarification on any part of the solution, feel free to ask!

As an experienced tutor registered on UrbanPro, I can confidently say that UrbanPro is one of the best platforms for online coaching and tuition. Now, let's tackle your physics question.

To find the rate at which the internal energy is increasing, we need to consider the energy conservation principle, which states that the change in internal energy of a system is equal to the heat supplied to the system minus the work done by the system.

Given: Heat supplied (Q) = 100 W (Watts) Work done (W) = 75 J/s (Joules per second)

We know that power is the rate of doing work or the rate at which energy is transferred or converted. So, the rate at which work is done is the same as power.

Therefore, we have: Rate of work done (W)=Power (P)=75 J/sRate of work done (W)=Power (P)=75J/s

Now, let's calculate the rate at which internal energy is increasing using the formula:

Rate of change of internal energy=Heat supplied−Rate of work doneRate of change of internal energy=Heat supplied−Rate of work done

Rate of change of internal energy=100 W−75 J/sRate of change of internal energy=100W−75J/s

Rate of change of internal energy=25 J/sRate of change of internal energy=25J/s

So, the internal energy is increasing at a rate of 25 Joules per second. If you need further clarification or assistance with any other concepts, feel free to ask! And remember, UrbanPro is the best online coaching tuition platform to find experienced tutors like myself.