GATE ME 1. Thermodynamics Chapter Thermodynamics Test

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Thermodynamics Test - 1

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Question 1
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A single fixed point temperature scale is based on

A
Ice point

B
Steam point

C
Triple point of water

D
Critical point of water

Solution

A single fixed point chosen as the basis of temperature scale is the Kelvin scale. The state in which ice, liquid water, and water vapor coexist in equilibrium, a state known as the triple point of water, provides the standard reference temperature. The temperature of the triple point of water, which can be very accurately and reproducibly measured, was assigned the value of 273.16 K, corresponding to 0.01°C, in order to maintain the magnitude of a unit temperature.
Question 2
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A
A

B
B

C
C

D
D

Solution

A heat engine is a system that converts heat into work by taking heat from the reservoir ( hot body) to carry out some work. There is a discharge of some heat to the sink (cold body).

Hot Air Balloon is used to lift the weight using heat.

A bimetallic strip is used to convert a temperature change into mechanical displacement. The strip consists of two strips of different metals which expand at different rates as they are heated. ... This effect is used in a range of mechanical and electrical devices.
Question 3
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A fan is used to deliver air at the rate of 5 m³/s with a velocity of 10 m/s. Find the power in Watts that must be supplied to the fan. Density of air = 1.2 kg/m³.

A
299.5-300.5

B
299.5-300.6

C
299.5-300.7

D
299.5-300.8

Solution

The power required by the fan will be equal to the rate of change of kinetic energy of the air which is caused by the fan.

The rate of change of kinetic energy of the air

dE = 1/2 × m˙×V²

where ṁ is the mass flow rate and V is the velocity of the air.

ṁ = ρ × Q

Calculation:

Given,ρ = 1.2 kg/m3, Q = 5 m3/s, V = 10 m/s

Assuming no heat loss in the fan, the power supplied to the fan

dE = 1/2 × m˙× V²

dE = 1/2 × 1.2 × 5 × 10²= 300 W
Question 4
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A cylinder contains 5 m³of ideal gas at a pressure of 1 bar. This gas is compressed in a reversible isothermal process till its pressure increases to 5 bar. The work in kJ required for this process is

A
804.7

B
953.2

C
981.7

D
1012.2

Solution

Work done in a reversible isothermal process is given by

W = C ln P₁/P₂= C ln V₂/V₁

Where C = P₁V₁ = P₂V₂

P is the pressure and V is the volume of the gas.

Calculation:

Given, P₁ = 1 bar, V₁ = 5 m³ and P₂ = 5 bar

C = P₁V₁ = 1 bar × 5 m³ = 100 kPa × 5 m³ = 500 kJ

W = 500 ln(1/5) = −804.718 kJ

-ve sign shows that work is done on the system.
Question 5
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One reversible heat engine operates between 1000 K and T₂K and another reversible heat engine operates between T₂K and 400 K. If both the engines have the same heat input and output, then the temperature T₂ must be equal to:

A
582.7 K

B
632.5 K

C
682.8 K

D
732.5 K

Solution
Question 6
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A window receives 200 W of heat transfer at the inside surface of 20°C and transmits the 200 W from its outside surface at 2°C continuing to ambient air at -5°C. Find the window’s rate of entropy generation.

A
0.68 W/K

B
0.73 W/K

C
0.045 W/K

D
0.064 W/K

Solution
Question 7
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Two air streams with mass flow rates of 36 kg/min and 14 kg/min with respective enthalpies of 36 kJ/kg da and 50 kJ/kg da are mixed. The enthalpy of the mixture is nearly

A
64 kJ/kg da

B
55 kJ/kg da

C
46 kJ/kg da

D
40 kJ/kg da

Solution

From the conservation of energy, Energy of the mixture = Sum of the energy of the individual streams.

The energy carried by flowing steam = mass flow rate × enthalpy/unit mass

Calculation:

Given, ṁ1 = 36 kg/min, ṁ2 = 14 kg/min, h₁ = 36 kJ/(kg da), h₂ = 50 kJ/(kg da)

Energy of the stream 1 = ṁ₁ × h₁ = 36 × 36 = 1296 kJ/(min da)

Energy of the stream 2 = ṁ₂ × h₂ = 14 × 50 = 700 kJ/(min da)

After the two streams mix into the one the resulting mass flow rate ṁ_mix = 36 + 14 = 50 kg/min

Now the energy of the resulting mass = h_mix × ṁf

From energy conservation E_mix = E₁ + E₂

⇒ h_f × 50 = 1296 + 700

⇒ h_mix= 1996/50 = 39.92 kJ/kg da
Question 8
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Heat and work are

A
intensive properties

B
extensive properties

C
point functions

D
path functions

Solution

Heat and work depend upon the path traversed to move from one thermodynamic state to another. This is why they are known as path functions.
Question 9
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The thermal efficiency of the hypothetical cycle shown is

A
0.6

B
0.5

C
0.4

D
0.3

Solution

The area under the curve on T-S diagram shows the heat interaction for a reversible process

The area enclosed by the cycle on the T-S diagram gives the work interaction for the cycle

In the cyclic process

W_net = Q_net = Q_In – Q_Out

Qout = Heat rejection at the lowest temperature in the cycle

Q_out = T_min∆S

Where ∆S is the entropy change

Calculation:

Net work is equal to the area enclosed by the cycle, i.e. area of the triangle
Question 10
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Which one of the following relationships defines the Helmholtz function F?

A
F = H + TS

B
F = H – TS

C
F = U – TS

D
F = U + TV

Solution

There are eight properties of a system, namely pressure (p), Volume (V), temperature (T), internal energy (u), enthalpy (h), entropy (s), Helmholtz function (f) and Gibbs function (g) and h, f and g are referred to as thermodynamic potentials.

Helmholtz function F = U - TS

Gibbs free function G = H - TS
Question 11
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A balloon containing an ideal gas is initially kept in an evacuated and insulated room. The balloon ruptures and the gas fills up the entire room. Which one of the following statements is TRUE at the end of above process?

A
The internal energy of the gas decreases from its initial value, but the enthalpy remains constant

B
The internal energy of the gas increases from its initial value, but the enthalpy remains constant

C
Both internal energy and enthalpy of the gas remain constant

D
Both internal energy and enthalpy of the gas increase

Solution

It is free expansion. Since vacuum does not offer any resistance, there is no work transfer involved in free expansion.
Question 12
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If the work done on a closed system is 20 kJ/kg, and 40 kJ/kg heat is rejected from the system, change in internal energy is

A
20 kJ/kg

B
60 kJ/kg

C
-20 kJ/kg

D
-60 kJ/kg

Solution

The first law of thermodynamics gives the relationship between the heat transfer work transfer and the change in energy associated with a process. It is given by

dQ = ΔU + dW

where dQ is the heat interaction, dW is the work interaction and ΔU is the change in internal energy of the system.

Sign convention for the Heat and work transfer

If heat is added to the system it is treated as + ve and heat rejected by the system is treated as –ve.

If work is added to the system it is treated as –ve and work done by the system is treated as +ve.

Calculation:

Given, dW = -20 kJ/kg (work is done on the system)

dQ = -40 kJ/kg (Heat is added to the system)

Then, from first law of thermodynamics

ΔU = dQ – dW

ΔU = -40 + 20 = -20 kJ/kg