Thermodynamics

We know here Water is flowing at a rate of 3.0 litre/min.

The geyser heats the water and raises its temperature from 27°C to 77°C. So change in temperature of water or Rise in temperature is

ΔT = 77°C – 27°C = 50°C

We know heat supplied to a substance in raising its temperature is given by relation

ΔQ = mc ΔT

Where ΔQ is the heat supplied to the substance of mass m Kg having specific heat c to raise its temperature by ΔT

We know Specific heat of water is

c = 4.2 J g^-1°C^-1

we will calculate for 1 minute so we will consider mass of water flowing every minute

Now volume of water flowing every minute

V = 3.0 litre/min

We know

Mass = volume × density

Density of water = 1000g/l

So we have mass of water flowing every minute

m = 3.0 litre/min × 1000g/l

i.e. m = 3000 g/min

putting values of m ,c , ΔT in the equation ΔQ = mc ΔT

we get

ΔQ = 3000g/l × 4.2 J g^-1°C^-1 × 50°C

= 6.3 × 10⁵ J/min

So per minute 6.3 × 10⁵ J heat is required by water

But heat of combustion is of gas is

ΔH = 4.0 × 10⁴ J/g

i.e. per gram of gas 4.0 × 10⁴ J heat is released

so the rate of consumption of gas will be

R = ΔQ/ΔH

Where ΔQ is heat consumed per minute and ΔH is heat supplied per gram of gas

So Rate of consumption will be

= 15.75 g/min

That is each minute 15.75 g fuel is being consumed

we know amount of heat transferred to any substance (solid , liquid, Gas) at constant pressure is given by the relation

ΔQ = nC_PΔT

Where ΔQ is the amount of heat supplied to the substance in raising it temperature by ΔT, C_P is the molar specific heat of substance at constant pressure, n is the number of moles of substance.

Here the Substance is Nitrogen Gas (N₂) , which is a diatomic gas, for a diatomic Gas the molar specific heat is

R is universal Gas Constant , R = 8.3 J mol^–1 K^–1

The rise in temperature is of 45°C, so we have

ΔT = 45°C = 45 K

(since we are only concerned with difference of temperature so unit can be either ^oC or K , both have same interpretation)

To find the moles of N₂ Gas we have the relation

n = m/M

where n is the number of moles of gas, m is the mass of gas in grams (g) and M is the Molar mass of the gas in g/mol

here we are given mass of N₂ as

m = 2.0 × 10^–2 kg

1 Kg = 1000 g , so we have

m = 2.0 × 10^–2 × 10³ = 20 g

molar mass of N₂ gas is

M = 28 g/mol

So the number of moles of N₂ gas is

so putting the value of ΔT,n,C_P in the equation ΔQ = nC_PΔT

ΔQ = 0.714mol ×(7/2) × R × 45 K

= 0.714mol × 3.5 × 8.3 J mol^–1 K^–1 × 45K

= 933.38 J

So 933.38 J heat must be supplied to 20g nitrogen gas to raise its temperature by 45^oC at constant Pressure

When Two bodies at different temperatures T₁ and T₂ if brought in thermal contact , heat flows from the body at higher temperature to the body at lower temperature , till both attain thermal equilibrium or we can say temperature of both become same and equal to the mean temperature (T₁ + T₂ )/2 . but this is not always the case if heat capacities of the both the bodies is not same. since we know amount of heat transferred to a body is proportional to its heat capacity , so if heat capacity of one body is less than that of other it will absorb less heat and temperature of both will not become equal and both will not settle to the mean temperature (T₁ + T₂ )/2

The coolant in a chemical or nuclear plant should have a high specific heat since amount of heat transferred to a body is proportional to its Specific heat, so if specific heat is higher it could absorb more heat itself thus preventing other parts of plant from getting too hot and itself consuming most of the excess heat produced in the chemical or nuclear Plant.

When a car is driven, wheels rotate and are in contact with ground and also experience frictional force, we know due to friction heat is also generated so tyres become hot or the temperature of tyres rises due to which temperature of gas inside the tyre also rises, but the volume of the tyre remains same , we know from Charles law when volume is constant , pressure is directly proportional to Temperature, since temperature is rising so pressure of tyre due to gas also rises.

Harbour Town is near to a Water Body due to which there is huge amount of humidity in air or we can water vapours are present in greater content in air as compared to a desert town at same latitude, so due to water vapour heat capacity of air in harbour town increases so it absorbs greater amount of heat during day time and also releases it slowly during night maintaining temperate temperature as compared to Desert region where there is very less humidity in air.

Now here the cylinder or the container containing the gas and piston mounted are both insulated so no transfer of heat will take place either from system to surrounding and vice versa , so the process will be adiabatic as the heat supplied to or from the system

ΔQ = 0

Now the system consists of 3 Mole Hydrogen gas kept at standard temperature and pressure in an insulated container

As shown in the figure

Now here the gas is compressed to half of its volume, we have to calculate factor of change of pressure of gas when volume is reduced to half

We know for an adiabatic process we have the relation

Where P₁ and P₂ are initial and final pressures of the system respectively, V₁ and V₂ are initial and final volume of the system respectively, γ is the specific heat ratio of the gas since, H₂ is a diatomic gas, for a diatomic gas

γ = 1.4

now let the initial volume of the system be

V₁ = V

Now here the gas is compressed to half of its volume so final volume of the system is

V₂ = V/2

Let initial and final pressures of the system be P₁ and P₂ respectively

Now applying the relation

We have

Or

i.e.

so we can say that the final pressure is 2.639 times initial pressure or the pressure increased by a factor of 2.639

We know in an adiabatic change there is no transfer of heat either from system to surrounding or from surrounding to system, so there is no change in total heat Energy of system i.e.

ΔQ = 0

Now amount of work equal to 22.3 J is done on the system so we have

ΔW = -22.3 J

(Negative sign states that work is done on the system)

We know The first law of thermodynamics is the general law of conservation of energy applied to any system in which energy transfer from or to the surroundings (through heat and work) and is stated as

ΔQ = ΔU + ΔW

where ΔQ is the heat supplied to the system, ΔW is the work done by the system and ΔU is the change in internal energy of the system.

So to find the change in internal energy of system we will apply the first law of thermodynamics to find change in internal Energy

ΔQ = 0 and ΔW = -22.3 J , ΔU = ?

0 = ΔU + (-22.3 J)

i.e. ΔU = 22.3 J

so change in internal energy of gas in adiabatically going from an equilibrium state A to another equilibrium state B is 22.3 J

now gas is again taken from equilibrium state A to equilibrium state B, since internal energy of a system is a state variable, this means it depends only upon initial and final state of the system irrespective of the type of process so here also change in internal energy would have same value

i.e. ΔU = 22.3 J

this time heat absorbed by the system is 9.35 cal

1 cal = 4.19 J

So ΔQ = 9.35 × 4.19 = 39.17 J

(ΔQ is positive because heat is absorbed by the system)

Now again using first law of thermodynamics to find the work done by the gas

ΔQ = ΔU + ΔW

where ΔQ is the heat supplied to the system, ΔW is the work done by the system and ΔU is the change in internal energy of the system.

Here

ΔQ = 39.17 J , ΔU = 22.3 J , ΔW = ?

So we have

39.17 J = 22.3 J + ΔW

i.e. ΔW = 39.17 – 22.3 J = 16.87 J

so the work done by the gas in this process is 16.87 J

Now here two insulated cylinders are connected to each other via a stop cock A contains a gas at standard temperature and pressure, and cylinder B is empty.

as shown in figure

Now as the cock is suddenly opened the gas from A will rush towards B, i.e. free expansion of gas will take place and the process will be very fast and not Quasi-static

(a) For free expansion of a gas we have the relation

P₁V₁ = P₂V₂

Where P₁ and P₂ are initial and final pressures of the system respectively, V₁ and V₂ are initial and final volume of the system respectively

Now we know initial system is container A only because gas is in A only but final system is Both container A and B combined and gas in also filled in container B uniformly

As shown in Figure

Now since initially system was at Standard Temperature and Pressure initial pressure of the system would be

P₁ = 1 atm

Let volume of both the cylinders be V

Since initially system was only container A so initial volume of the system would be

V₁ = V

We have to find final pressure which would be same in both the cylinders, let it be P₂

Now finally gas is distributed in both the cylinders so final volume will be the total volume of both the cylinders i.e.

V₂ = 2V

So applying

P₁V₁ = P₂V₂

We have

1atm × V = P₂ × 2V

or P₂ = 1/2 atm = 0.5 atm

so the final pressure in both the cylinders would be half of initial pressure in cylinder A and will be 0.5 atm

(b) now in free expansion no work is done by gas and system is already insulated so no heat transfer has taken place to or from the system, for internal energy of a gas to change either work should be done by or on the gas or heat should be transferred to or from the gas , but here none has happened so internal energy would remain same or constant or change in internal energy of gas is 0, i.e.

ΔU = 0

(c) In case of free expansion of gas no heat is transferred to or from the system, no work is done i.e. no change of internal energy of gas and temperature of a gas changes when its internal energy changes so the temperature remain constant, i.e. no change in temperature or we can say the change in temperature is 0, i.e.

ΔT = 0

(d) Free expansion is a very fast process , and not Quasi-static so no stable intermediate equilibrium states exist and hence the intermediate states do not satisfy the gas equation, they do not lie on the P-V-T surface of the system.

The steam engine absorbs energy in form of heat from boiler and converts some of it into mechanical work and return rest to the surrounding or we can say some part of heat absorbed is converted into work and rest is wasted, suppose Q₁ heat is absorbed from Boiler (Source), W amount of work is done by the engine and Q₂ heat is returned to the surroundings (Sink)

As has been shown diagrammatically

So we have

Q₁ = W + Q₂

Now since engine is absorbing 3.6 × 10⁹J of heat per minute from its boiler so each minute amount of heat absorbed by engine from Boiler is

Q₁ = 3.6 × 10⁹J

steam engine delivers 5.4×10⁸J of work per minute, so each minute work done by engine is

W = 5.4×10⁸J

Now efficiency is defined as the amount of work done per unit supplies Energy , or it is the ratio of work output and the heat input of the system , mathematically

Where W is the work done by engine and Q₁ is the amount heat absorbed

So efficiency of the engine is

In percentage form

So the efficiency of the engine is 15%

Now we need to find the amount of heat wasted, which is the amount of heat or energy transferred back to Surrounding (Sink) without doing any work, denoted by Q₂

We already know

Q₁ = W + Q₂

Or we can say

Q₂ = Q₁ – W

Now since engine is absorbing 3.6 × 10⁹J of heat per minute from its boiler so each minute amount of heat absorbed by engine from Boiler is

Q₁ = 3.6 × 10⁹J

steam engine delivers 5.4×10⁸J of work per minute, so each minute work done by engine is

W = 5.4×10⁸J

So we get

Q₂ = 3.6 × 10⁹J – 5.4 × 10⁸J

= 36 × 10⁸ J - 5.4 × 10⁸J = 30.6 × 10⁸J

=3.06 ×10⁹ J

i.e. 3.06 ×10⁹ J of heat is wasted each minute or we can say

3.06 ×10⁹ J heat per minute is being wasted

We know The first law of thermodynamics is the general law of conservation of energy applied to any system in which energy transfer from or to the surroundings (through heat and work) and is stated as

ΔQ = ΔU + ΔW

where ΔQ is the heat supplied to the system, ΔW is the work done by the system and ΔU is the change in internal energy of the system.

Here electric heater supplies heat to a system at a rate of 100W i.e. each second 100 J heat is supplied to the system so the heat supplied to system each second is

ΔQ = 100 J

system performs work at a rate of 75 joules per second, so work done by the system each second is

ΔW = 75 J

So putting in equation ΔQ = ΔU + ΔW

100 J = ΔU + 75 J

Or we get

ΔU = 100 J – 75 J = 25 J

So change in internal energy each second is 25 J , or we can say rate of change or increase of internal energy of system is 25 J/s

For any thermodynamic process represented by a curve with x axis representing volume and y axis representing Pressure, area under the curve between two points gives the work done in the process, further if volume is increasing the work done is positive , if volume is decreasing the work done is negative.

Now as we can see along DE the volume increased from 2.0m³ to

5.0m³ so work done along DE is positive and area under DE with x axis is the magnitude

As shown in figure

Now as we can see along EF the volume decreased from 5.0m³ to

2.0m³ so work done by the gas along EF is negative and area under EF with x axis is the magnitude

As shown in figure

Positive and negative work cancel each other so the net magnitude of work done by gas is the area of triangle DEF and magnitude will be positive

As shown in figure

Now to find the area we know area of a triangle is given by

A = 1/2 × Base × Height

Here Base of triangle is EF, Height is DF

EF = (5.0m³ – 2.0m³) = 3.0 m³

FD = (600Nm^-2 – 300Nm^-2) = 300Nm^-2

So the work done is

W = 1/2 × EF × FD

= 1/2 × 3.0 m³ × 300Nm^-2

= 450 J

So the total work done by the gas from D to E to F in the thermodynamic process is 450 J

A refrigerator is a device which is maintained at some low temperature by transferring heat to surrounding at high temperature We know coefficient of performance of a refrigerator in terms of temperature is given by the relation

Where is the coefficient of Performance , T₂ is temperature of surrounding , T₁ is the temperature at which refrigerator is maintained

we are given temperature of refrigerator

T₁ = 9 ^oC

Converting it to Kelvin

T₁ = 9 + 273 = 282 K

and we are given temperature of surrounding

T₂ = 36 ^oC

Converting it to Kelvin

T₂ = 36 + 273 = 309 K

So we the coefficient of performance is

= 10.44

So the coefficient of performance of refrigerator is 10.44