A flat-plate solar collector has a black absorber plate maintained at 80°C. The ambient temperature is 20°C and the heat loss coefficient (overall U value including radiation and convection) is 8 W/(m²·K). For a collector area of 4 m², what is the he

The correct answer is C: 1920 W. Q_loss = U × A × ΔT = 8 × 4 × (80-20) = 8 × 4 × 60 = 1920 W. The overall U value already accounts for both convective and radiative losses.

For radiation heat exchange between two parallel plates at temperatures T₁ and T₂ with emissivities ε₁ and ε₂, the radiation heat transfer is reduced by factor:

The correct answer is B: F = ε₁ε₂/(ε₁ + ε₂ - ε₁ε₂). For two parallel plates, Q = σA·F(T₁⁴ - T₂⁴) where F = ε₁ε₂/(ε₁ + ε₂ - ε₁ε₂), derived from network resistance analogy.

The enthalpy change for a constant pressure process equals:

The correct answer is A: Heat absorbed by the system. At constant pressure, ΔH = qₚ (heat at constant pressure). This is the definition of enthalpy and its primary application.

For a reversible process at constant T and P, the minimum work required (excluding PV work) is:

The correct answer is B: w_useful = -ΔG. Useful work (non-PV) available = -ΔG at constant T,P. This represents maximum useful work for spontaneous process or minimum work needed for non-spontaneous process.

For a polymerization reaction in a batch reactor, the polydispersity index (PDI) and average molecular weight depend on:

The correct answer is B: The ratio of propagation to termination rate constants and conversion. PDI and molecular weight distribution in batch polymerization depend on the relative magnitudes of propagation and termination reactions and the degree of conversion achieved.

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Chemical Engineering
Thermodynamics

Process design, thermodynamics, reactions

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All Mass Transfer 100 Heat Transfer 100 Fluid Mechanics 100 Chemical Reaction Engineering 100 Thermodynamics 97

Q.1 Easy Thermodynamics

A system absorbs 500 J of heat and does 300 J of work. The change in internal energy is:

A 200 J

B 800 J

C -200 J

D 500 J

Correct Answer: A. 200 J

EXPLANATION

First law: ΔU = Q - W = 500 - 300 = 200 J (using convention W = work by system).

Test

Q.2 Easy Thermodynamics

What does the Gibbs free energy criterion ΔG < 0 indicate for a process?

A Process is non-spontaneous at all temperatures

B Process is spontaneous and irreversible

C Process is at equilibrium

D Process releases maximum useful work

Correct Answer: B. Process is spontaneous and irreversible

EXPLANATION

ΔG < 0 indicates spontaneous, irreversible process under constant T and P conditions.

Test

Q.3 Easy Thermodynamics

For a system at equilibrium, the chemical potential of a substance in different phases must be:

A Equal

B Different

C Zero

D Proportional to temperature

Correct Answer: A. Equal

EXPLANATION

At phase equilibrium, μ_liquid = μ_vapor = μ_solid. This equality determines equilibrium conditions.

Test

Q.4 Easy Thermodynamics

What is the Clausius-Clapeyron equation used for?

A Calculating work done in cyclic processes

B Relating vapor pressure to temperature during phase transitions

C Determining heat capacity of substances

D Calculating entropy of mixing

Correct Answer: B. Relating vapor pressure to temperature during phase transitions

EXPLANATION

Clausius-Clapeyron equation: ln(P₂/P₁) = -(ΔH_vap/R)(1/T₂ - 1/T₁) describes phase equilibrium.

Test

Q.5 Easy Thermodynamics

In the van der Waals equation, the term 'a' represents:

A Volume correction due to molecular size

B Pressure correction due to intermolecular attractive forces

C Temperature correction factor

D Compressibility factor at critical point

Correct Answer: B. Pressure correction due to intermolecular attractive forces

EXPLANATION

In (P + a/V²)(V - b) = RT, 'a' corrects for intermolecular forces reducing pressure, while 'b' corrects for molecular volume.

Test

Q.6 Easy Thermodynamics

A gas undergoes an isothermal expansion from 2 L to 5 L at 298 K. If the process is reversible, what is the sign of entropy change for an ideal gas?

A Positive

B Negative

C Zero

D Cannot be determined

Correct Answer: A. Positive

EXPLANATION

For isothermal expansion of an ideal gas, ΔS = nR ln(V_f/V_i) = nR ln(5/2) > 0. Volume increases, so entropy increases.

Test

Q.7 Easy Thermodynamics

During a constant volume process, the first law of thermodynamics simplifies to:

A ΔU = q + w = q - PΔV

B ΔU = q (since w = 0 at constant volume)

C ΔU = w (since q = 0 in all processes)

D ΔU = 0 (no energy change)

Correct Answer: B. ΔU = q (since w = 0 at constant volume)

EXPLANATION

At constant volume, ΔV = 0, so w = -P∫dV = 0. Therefore, ΔU = q + w = q + 0 = q. All heat goes into internal energy change.

Test

Q.8 Easy Thermodynamics

For a binary ideal solution, the total vapor pressure at constant T is given by:

A P = P₁°x₁ + P₂°x₂ (Raoult's law)

B P = P₁°/x₁ + P₂°/x₂

C P = (P₁° + P₂°)/(x₁ + x₂)

D P = P₁° + P₂° regardless of composition

Correct Answer: A. P = P₁°x₁ + P₂°x₂ (Raoult's law)

EXPLANATION

Raoult's law states P_i = P_i°x_i for ideal solutions. Total pressure P = P₁°x₁ + P₂°x₂. This assumes ideal mixing behavior.

Test

Q.9 Easy Thermodynamics

In a constant pressure process, the heat absorbed by a system equals:

A Change in internal energy only

B Enthalpy change

C Work done by the system

D Change in entropy

Correct Answer: B. Enthalpy change

EXPLANATION

At constant pressure, q_p = ΔH (change in enthalpy). This is the definition of enthalpy and is a key relationship in engineering thermodynamics.

Test

Q.10 Easy Thermodynamics

For an ideal gas undergoing adiabatic compression, the entropy change is:

A Positive

B Negative

C Zero

D Indeterminate without volume ratio

Correct Answer: C. Zero

EXPLANATION

For a reversible adiabatic process, dq = 0, therefore ΔS = ∫dq_rev/T = 0. Entropy remains constant during reversible adiabatic processes.

Test

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