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Correct option 1. The 2^nd Law

Explanation:

The second law of thermodynamics states that the entropy of the universe is always increasing. Consequently, only processes which increases the overall entropy of the universe satisfy the second law and happen spontaneously.

Correct option (4) water

Explanation:

Water has the highest specific heat of the substances listed, so it has the capacity to emit the largest quantity of heat with minimal temperature loss; the emitted heat keeps you warm. The substance continues to warm you until its temperature is at or below your body temperature.

Correct option (6) −15.2 kJ/mol

Explanation:

We use Hess’ Law:

∆H◦ = ∑ n ∆H◦ _j,prod − ∑ n ∆H◦ _j,reac

= ∆H◦ _{f, CaCl2(aq)} + ∆H◦ _{f, H2O(ℓ)}

+ ∆H◦ _f,CO2(g) − [ ∆H◦ _{f, CaCO3(s)} 2 + (∆H◦ _{f, HCl(aq)}]

= −877.1 kJ/mol + (−285.83 kJ/mol)

+ (−393.51 kJ/mol) − [ −1206.9 kJ/mol + 2 (−167.16 kJ/mol)]

= −15.22 kJ/mol .

A volume of 1 L equals 0.001 m3, 

 the density is in the range of 10000 kg/m³ 

so we get

m = ρV = 10 000 kg/m³ × 0.001 m³ = 10 kg

A more accurate value  is ρ

= 13580 kg/m³. 

For the air  1 kg/m³ so we get

m = ρV = 1 kg/m³ × 0.001 m³ = 0.001 kg

 A more accurate value is ρ

= 1.17 kg/m³ at 100 kPa, 25°C. 

The reversible work is given by

if no kinetic or potential energy changes

w = −∫ v dP

The liquid has a very small value for v compared to a large value for a gas. 

If the process is said to be adiabatic then:

Steady state adiabatic single flow :

s_e = s_i + s_gen ≥ s_i

let us follow the water flow. It starts out with kinetic and potential energy of some magnitude at a compressed liquid state P, T. As the water splashes onto the car it looses its kinetic energy (it turns in to internal energy so T goes up by a very small amount). As it drops to the ground it then looses all the potential energy which goes into internal energy. Both of theses processes are irreversible so s goes up.

If the water has a temperature different from the ambient then there will also be some heat transfer to or from the water which will affect both T and s. 

(C) U₁ increases and U₂ decreases and the total entropy increases

Heat is energy transferred across the boundary of a system due to temperature difference between the system and the surrounding. The heat can be transferred by conduction, convection and radiation. The main characteristics of heat are:

1. Heat flows from a system at a higher temperature to a system at a lower temperature. 

2. The heat exists only during transfer into or out of a system. 

3. Heat is positive when it flows into the system and negative when it flows out of the system. 

4. Heat is a path function. 

5. It is not the property of the system because it does not represent an exact differential dQ. It is therefore represented as δQ.

Heat required to raise the temperature of a body or system, Q = mc (T₂ – T₁)

Where, m = mass, kg

T₁, T₂ = Temperatures in°C or K

c = specific heat, kJ/kg–K

Specific heat for gases can be specific heat at constant pressure (C_p) and constant volume (c_v)

Also; mc = thermal or heat capacity, kJ.

mc = water equivalent, kg.

WORK 

The work may be defined as follows: “Work is defined as the energy transferred (without transfer of mass) across the boundary of a system because of an intensive property difference other than temperature that exists between the system and surrounding.” Pressure difference results in mechanical work and electrical potential difference results in electrical work.

Or

“Work is said to be done by a system during a given operation if the sole effect of the system on things external to the system (surroundings) can be reduced to the raising of a weight”. The work is positive when done by the system and negative if work is done on the system

Hg : L = 1 m; ρ = 13 580 kg/m³ 

The pressure difference ∆P balances the column of height L 

∆P = ρ g L = 13 580 kg/m³ × 9.80665 m/s² × 1.0 m × 10^-3 kPa/Pa

= 133.2 kPa

The lowest temperature is absolute zero which is at zero degrees Rankine at which point the temperature in Fahrenheit is negative

T_R = 0 R = −459.67 F 

No.The exergy equation is derived from the other blance equations by defining the exergy from the state properties and the reference dead state. 

No. You decrease its availability by bringing it closer to To, where it has zero availability, if we neglect pressure effects. Any substance at a T different from ambient (higher or lower) has a positive availability since you can run a heat engine using the two temperatures as the hot and cold reservoir, respectively. For a T lower than the ambient it means that the ambient is the hot side of the heat engine. 

Yes. They are very similar. Reversible work is usually defined as the reversible work that can be obtained between two states, inlet-exit or beginning to end. Availability is a property of a given state and defined as the reversible work that can be obtained by changing the state of the substance from the given state to the dead state (ambient).  

No. By virtue of its definition kinetic energy can only be positive. The potential energy is measured from a reference elevation (standard sea level or a local elevation) so it can be negative. The thermodynamic state can only have a positive exergy the smallest it can be is zero if it is the ambient dead state. 

Since the wind is generated by a complex system driven by solar heat input and radiation out to space it is a kind of heat engine. 

Within our lifetime it looks like it is perpetual. However with a different time scale the climate will change, the sun will grow to engulf the earth as it burns out of fuel. 

Given that P_abs = 30mm of Hg 

P_abs = ρgh = 13550 × 9.78 × 30/1000 = 39.755 × 10³ N/m²                           ...(i) 

P_atm =ρgh = 13550 × 9.78 × 760/1000 = 100714.44 N/m²                             ...(ii)

P_gauge = P_abs – P_atm 

= 39.755 × 10³ – 100714.44

= – 60958.74 N/m²                                            ......ANS

Thermodynamics is the science that deals with the conversion of heat into mechanical energy. It is based upon observations of common experience, which have been formulated into thermodynamic laws. These laws govern the principles of energy conversion. The applications of the thermodynamic laws and principles are found in all fields of energy technology, notably in steam and nuclear power plants, internal combustion engines, gas turbines, air conditioning, refrigeration, gas dynamics, jet propulsion, compressors, chemical process plants, and direct energy conversion devices. 

Generally thermodynamics contains four laws; 

1. Zeroth law: deals with thermal equilibrium and establishes a concept of temperature. 

2. The First law: throws light on concept of internal energy. 

3. The Second law: indicates the limit of converting heat into work and introduces the principle of increase of entropy. 

4. Third law: defines the absolute zero of entropy. 

These laws are based on experimental observations and have no mathematical proof. Like all physical laws, these laws are based on logical reasoning.

 Thermodynamics is the study of energy, energy and entropy. 

The whole of heat energy cannot be converted into mechanical energy by a machine. Some portion of heat at low temperature has to be rejected to the environment. 

The portion of heat energy, which is not available for conversion into work, is measured by entropy.

 The part of heat, which is available for conversion into work, is called energy.

 Thus, thermodynamics is the science, which computes energy, energy and entropy.

Given that h = 760mm of Hg for P_atm 

P_atm = ρgh = 13.6 × 10³ × 9.81 × 760/1000 = 101396.16

 N/m² = 101.39 × 10³ N/m²                                                                                            ...(i)

 P_gauge = ρgh = 1000 × 9.81 × 10/100 = 981 N/m₂                                         ...(ii) 

P_abs = P_atm + P_gauge

 = 101.39 × 10³ + 981

 P_abs = 102.37×10³ N/m²      .......ANS

For defining any system certain parameters are needed. Properties are those observable characteristics of the system, which can be used for defining it. For example pressure, temp, volume.

 Properties further divided into three parts;

Intensive Properties 

Intensive properties are those, which have same value for any part of the system or the properties that are independent of the mass of the system. EX; pressure, temp.

 Extensive Properties

 Extensive properties are those, which dependent Upon the mass of the system and do not maintain the same value for any part of the system. EX; mass, volume, energy, entropy.

 Specific Properties

 The extensive properties when estimated on the unit mass basis result in intensive property, which is also known as specific property.

 EX; sp. Heat, sp. Volume, sp. Enthalpy.

Correct option 4. The reaction is exothermic

Explanation:

Since the reaction becomes more and more spontaneous as the temperature is lowered, it must be spontaneous at T = 0 K. Since ∆G = ∆H at T = 0 K, ∆H must be negative and the reaction is exothermic.

Correct option (6) III only 

Explanation:

Examples of extensive state functions include: mass, volume, enthalpy, entropy, internal energy, free energy etc.

Thermodynamic studies are undertaken by the following two different approaches.  Macroscopic approach–(Macro mean big or total)

 Microscopic approach–(Micro means small)

 The state or condition of the system can be completely described by measured values of pressure, temperature and volume which are called macroscopic or time–averaged variables. In the classical

thermodynamics, macroscopic approach is followed. The results obtained are of sufficient accuracy and validity. Statistical thermodynamics adopts microscopic approach. It is based on kinetic theory. The matter consists of a large number of molecules, which move, randomly in chaotic fashion. At a particular moment, each molecule has a definite position, velocity and energy. The characteristics change very frequently due to collision between molecules. The overall behaviour of the matter is predicted by statistically averaging the behaviour of individual molecules. 

Microscopic view helps to gain deeper understanding of the laws of thermodynamics. However, it is rather complex, cumbersome and time consuming. Engineering thermodynamic analysis is macroscopic and most of the analysis is made by it. 

These approaches are discussed (in a comparative way) below:

Newtons 2^nd law: F = ma

F = ma = 2 lbm × 5 ft/s² = 10 lbm ft/s² 

= 10/32.174 Ibf = 0.31 Ibf

A volume of 1 gal equals 231 in³, see

 the density is in the range of 10 000 kg/m³ = 624.28 lbm/ft³, so we get 

m = ρV = 624.3 lbm/ft³ × 0.25 × (231/12³) ft³ = 20.86 lbm

A more accurate value ρ = 848 lbm/ft³.

For the air that density is about 1 kg/m³ = 0.06243 lbm/ft³ so we get

m = ρV = 0.06243 lbm/ft³ × 0.25 × (231/12³) ft³ = 0.00209 lbm

A more accurate value is ρ = 0.073 lbm/ft³ at 77 F, 1 atm. 

Thermodynamics has very wide applications as basis of thermal engineering. Almost all process and engineering industries, agriculture, transport, commercial and domestic activities use thermal engineering. But energy technology and power sector are fully dependent on the laws of thermodynamics. 

For example: 

(i) Central thermal power plants, captive power plants based on coal.

 (ii) Nuclear power plants. 

(iii) Gas turbine power plants. 

(iv) Engines for automobiles, ships, airways, spacecrafts. 

(v) Direct energy conversion devices: Fuel cells, thermoionic, thermoelectric engines. 

(vi) Air conditioning, heating, cooling, ventilation plants. 

(vii) Domestic, commercial and industrial lighting. 

(viii) Agricultural, transport and industrial machines. All the above engines and power consuming plants are designed using laws of thermodynamics.

Take a C.V. around the whole kitchen. The only energy term that crosses the control surface is the work input W. apart from energy exchanged with the kitchen surroundings. That is the kitchen is being heated with a rate of W. 

Remark: The two heat transfer rates are both internal to the kitchen. Q. H goes into the kitchen air and Q. L actually leaks from the kitchen into the refrigerated space, which is the reason we need to drive it out again. 

The valve and the cold line, the evaporator, is inside close to the inside wall and usually a small blower distributes cold air from the freezer box to the refrigerator room.

The black grille in the back or at the bottom is the condenser that gives heat to the room air. The compressor sits at the bottom. 

Given that

State 1 : 10 bar 280°C

State 2 : 1.6 bar, 0.8 dry

Specific heat of superheated steam = 2.1 kJ/kg K

Internal energy at state 1 is:

u₁ = u_g + m.c.(T₁ – T_sat) = (h_g - pv_g) + m.c. (T₁ - T_sat)

= (2776.2 – 1000 × 0. 19429) + 2.1(280 – 179.88) = 2792.16 kJ/kg

Internal energy at state 2;

u₂ = uf₂ + xu_f g₂

= (h_f – Pv_f )₂ + x [h_fg – P (v_fg)]₂

= (h_f – Pv_f ) + x (h_g – h_f) – P (v_g – v_f )]

= (h_f – Pv_f ) + x ((h_g – Pvg) – (h_f – P_vf))

= [475.38 + 160 (0.0010547)] + [(2696.2 – 160 * (1.0911))

– (475.38 – 160 (0.0010547))]

= 475.21 + 0.8 (2521.62 – 475.21)

= 2112.34 kJ/kg

Change in internal energy = 211234 - 2792.16 = - 679.82 kJ/kg

-ve sign shows the reduction in internal energy

Given data: 

V = 50Km/h = 50 × 5/18 = 13.88 m/sec

F = 900N

Power = ?

Q = ?

P = F.V = 900 × 13.88 = 12500 W = 12.5KW 

Heat equivalent of W.D. per minute by the engine = power × 1 minute

= 12.5 KJ/sec × 60 sec = 750KJ

 Q =750KJ

The difference in the level of the two limbs = P_gauge

 P_gauge = P_abs – P_atm

P_abs – P_atm = 562mm of Hg

P_abs – 101.39 = ρgh = 13.6 × 10³ × 9.81 × 562/1000 = 75.2 × 10³ N/m² 

= 75.2KPa

P_abs = 101.39 + 75.2 = 176.5kPa

 ANS: P = 176.5kPa

(a) Fuel and air enter, warm products of the combustion exit, large -Q to the air in the duct system, small -Q loss directly to the room. 

(b) Fuel and air enter, warm products exit through the chimney, cool air into the cold air return duct, warm air exit hot-air duct to heat the house. Small heat transfer losses from furnace, chimney and ductwork to the house. 

ρ = 1008 – TC/2 [kg/m³]

We need to express degrees Celsius in degrees Kelvin

T_C = T_K – 273.15

and substitute into formula

ρ = 1008 – T_C/2 = 1008 – (T_K – 273.15)/2 = 1144.6 – T_K/2

W =  ∫Fdx = F ∫dx = F Δx

= 200 × 30 = 6000 ft lbf

= (6000/778) Btu = 7.71 Btu

W = W / Δt = 7.71 / 60

= 0.129 Btu/s

For a complete cycle the substance has no change in energy and therefore no storage,

so the net change in energy is zero.

For a complete cycle the substance returns to its beginning state, so it has no change in specific volume and therefore no change in total volume.

1 cal (Int.) = 4.1868 J

= 4.1868 Nm = 4.1868 kg m²/s²

This was historically defined as the heat transfer needed to bring 1 g of liquid water from 14.5°C to 15.5°C, notice the value of the heat capacity of water

1 N-m = 1 J or Force times displacement = energy = Joule

The correct option is b

The heat of formation of a substance is the enthalpy associated with ONE mole of a substance from its elements in their standard states under standard conditions.

C_(graphite) + 2H₂ → CH₄(g)

Δ_rxn = enthalpy of formation of methane

In both processes a flow moves from a higher to a lower pressure. In the nozzle the pressure drop generates kinetic energy, whereas that does not take place in the throttle process. The pressure drop in the throttle is due to a flow restriction and represents a loss. 

Yes. By definition work is 100% exergy or availability

By definition the possible amount of work that can be obtained equals the exergy (availability). The maximum is limited to that out of a reversible heat engine, if constant T then that is the Carnot heat engine

W = (1 − To/T)Q

So we get a maximum for an infinite high temperature T, where we approach an efficiency of one. In practice you do not have such a source (the closest would be solar radiation) and secondly no material could contain matter at very high T so a cycle process can proceed (the closest would be a plasma suspended by a magnetic field as in a tokamak).  

P_sat  = 857.5 kPa at 20°C so superheated vapor.

v = 1.4153 m³/kg under subheading 100 kPa

m = V/v = 0.001 m³/1.41153 m³/kg = 0.000 706 kg = 0.706 g   

The boundary work is: W = ⌡P dV

P drops but does V go up or down?

The process equation is: PVⁿ = C

so we can solve for P to show it in a P-V diagram

P = CV^-n 

as n = 1.667 the curve drops as V goes up we see 

 V₂ > V₁ giving dV > 0 and the work is then positive. 

STATE 

The State of a system is its condition or configuration described in sufficient detail. State is the condition of the system identified by thermodynamic properties such as pressure, volume, temperature, etc. The number of properties required to describe a system depends upon the nature of the system. However each property has a single value at each state. Each state can be represented by a point on a graph with any two properties as coordinates. 

Any operation in which one or more of properties of a system change is called a change of state. 

Point Function

 A point function is a single valued function that always possesses a single – value is all states. For example each of the thermodynamics properties has a single – value in equilibrium and other states. These properties are called point function or state function.

 Or

 when two properties locate a point on the graph ( coordinates axes) then those properties are called as point function.

 For example pressure, volume, temperature, entropy, enthalpy, internal energy. 

Path Function

Those properties, which cannot be located on a graph by a point but are given by the area or show on the graph.

A path function is different from a point function. It depends on the nature of the process that can follow different paths between the same states. For example work, heat, heat transfer.

I. C.V. Food. 

(a) short term.: -Q from warm food to cold refrigerator air. Food cools.

(b) Long term: -Q goes to zero after food has reached refrigerator T.

II.  C.V. refrigerator space, not food, not refrigerator system 

(a) short term: +Q from the warm food, +Q from heat leak from room into cold space. -Q (sum of both) to refrigeration system. If not equal the refrigerator space initially warms slightly and then cools down to preset T. 

(b) long term:  small -Q heat leak balanced by -Q to refrigeration system.

Note: For refrigeration system CV any Q in from refrigerator space plus electrical W input to operate system, sum of which is Q rejected to the room. 

There are many similarities between heat and work. 

1. The heat and work are both transient phenomena. The systems do not possess heat or work. When a system undergoes a change, heat transfer or work done may occur. 

2. The heat and work are boundary phenomena. They are observed at the boundary of the system. 

3. The heat and work represent the energy crossing the boundary of the system. 

4. The heat and work are path functions and hence they are inexact differentials. 

5. Heat and work are not the properties of the system. 

6. Heat transfer is the energy interaction due to temperature difference only. All other energy interactions may be called work transfer. 

7. The magnitude of heat transfer or work transfer depends upon the path followed by the system during change of state.