Thermodynamics Quiz: Interpolating Property Tables

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Interpolating Property TablesQuestion 1 of 9

For ammonia refrigerant at 25°C, the saturation pressure is 1.003 MPa with hf = 298.9 kJ/kg and hfg = 1166.9 kJ/kg. At 30°C, Psat = 1.195 MPa with hf = 323.1 kJ/kg and hfg = 1145.1 kJ/kg. What is the approximate enthalpy of saturated liquid ammonia at 27°C?

305.4 kJ/kg

309.2 kJ/kg

313.8 kJ/kg

318.5 kJ/kg

322.9 kJ/kg

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Thermodynamics Quiz

Thermodynamics Quiz: Interpolating Property Tables

Practice Interpolating Property Tables in Thermodynamics with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Interpolating Property Tables, giving you a quick way to practice the rules, question types, and explanations that matter most for Thermodynamics.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

305.4 kJ/kg
309.2 kJ/kg (correct answer)
313.8 kJ/kg
318.5 kJ/kg
322.9 kJ/kg

Explanation: When you encounter saturated property data at discrete temperatures and need values at intermediate temperatures, you're dealing with interpolation – a fundamental skill in thermodynamics where exact tabulated values aren't available. For saturated liquid enthalpy at 27°C, you need to interpolate between the given data points. Using linear interpolation between 25°C and 30°C: At 25°C:

h_f = 298.9 \text{ kJ/kg}

At 30°C:

h_f = 323.1 \text{ kJ/kg}

The interpolation formula is:

h_f(27°C) = h_f(25°C) + \frac{27-25}{30-25} \times [h_f(30°C) - h_f(25°C)]

h_f(27°C) = 298.9 + \frac{2}{5} \times (323.1 - 298.9) = 298.9 + 0.4 \times 24.2 = 309.6 \text{ kJ/kg}

This matches closest to B) 309.2 kJ/kg. A) 305.4 kJ/kg represents an error where you might have used incorrect temperature ratios or made arithmetic mistakes in the interpolation calculation. C) 313.8 kJ/kg suggests you may have used a 3:2 ratio instead of 2:5, incorrectly weighting the interpolation toward the higher temperature. D) 318.5 kJ/kg indicates possible confusion about which temperature data to use or applying an incorrect interpolation method entirely. Study tip: Always set up interpolation systematically: identify your two boundary points, calculate the fraction of the interval you've moved, then apply that fraction to the property difference. Practice this method until it becomes automatic.

Question 2

A refrigerant R-134a system operates between pressures of 0.8 MPa and 1.2 MPa. If the enthalpy at the lower pressure saturated liquid state is 95.47 kJ/kg and at the higher pressure it is 117.8 kJ/kg, what is the approximate enthalpy at 1.0 MPa saturated liquid state?

104.2 kJ/kg
106.6 kJ/kg (correct answer)
108.1 kJ/kg
110.4 kJ/kg
112.7 kJ/kg

Explanation: When you encounter thermodynamic property values at different pressures, you're often dealing with interpolation between known states. This question tests your ability to estimate properties at intermediate conditions using the data provided. To find the enthalpy at 1.0 MPa, you can use linear interpolation since 1.0 MPa falls exactly halfway between the given pressures. With enthalpy at 0.8 MPa being 95.47 kJ/kg and at 1.2 MPa being 117.8 kJ/kg, the pressure difference is 0.4 MPa total, and 1.0 MPa is 0.2 MPa above the lower pressure. Using linear interpolation:

h_{1.0} = h_{0.8} + \frac{(P_{1.0} - P_{0.8})}{(P_{1.2} - P_{0.8})} \times (h_{1.2} - h_{0.8})

h_{1.0} = 95.47 + \frac{0.2}{0.4} \times (117.8 - 95.47) = 95.47 + 0.5 \times 22.33 = 106.635 \text{ kJ/kg}

This gives us approximately 106.6 kJ/kg, which is answer B. Answer A (104.2 kJ/kg) underestimates the value, possibly from incorrectly weighting the interpolation toward the lower pressure. Answer C (108.1 kJ/kg) and D (110.4 kJ/kg) both overestimate, potentially from calculation errors or incorrect interpolation ratios. Remember that for saturated liquid properties, linear interpolation with pressure typically provides good approximations over small pressure ranges. Always check that your interpolated value falls logically between the given boundary values.

Question 3

Refrigerant R-134a at -10°C exists as a two-phase mixture with 25% liquid by mass. At this temperature: Psat = 0.201 MPa, vf = 0.000793 m³/kg, vg = 0.09933 m³/kg. What is the specific volume of this mixture?

0.0251 m³/kg
0.0495 m³/kg
0.0745 m³/kg (correct answer)
0.0870 m³/kg
0.0995 m³/kg

Explanation: When you encounter a two-phase mixture problem in thermodynamics, you're dealing with a substance that exists partially as liquid and partially as vapor at saturation conditions. The key is using the quality-weighted average formula for specific volume. Since this mixture is 25% liquid by mass, the quality (vapor mass fraction) is x = 0.75. For any intensive property in a two-phase mixture, you calculate the weighted average using:

v = v_f + x(v_g - v_f)

Substituting the given values:

v = 0.000793 + 0.75(0.09933 - 0.000793) = 0.000793 + 0.75(0.098537) = 0.000793 + 0.073903 = 0.0747 \text{ m}^3/\text{kg}

This rounds to 0.0745 m³/kg, confirming answer C. Let's examine why the other options are incorrect. Answer A (0.0251 m³/kg) likely results from using x = 0.25 instead of 0.75 – confusing the liquid fraction with the vapor quality. Answer B (0.0495 m³/kg) appears to be a simple arithmetic average of vf and vg, ignoring the mass fractions entirely. Answer D (0.0870 m³/kg) is suspiciously close to vg, suggesting someone might have approximated by assuming the mixture behaves mostly like vapor. Remember this key strategy: when given liquid percentage, convert to vapor quality (x = 1 - liquid fraction) before applying the mixture property formula. Always double-check that your calculated value falls logically between the pure liquid and pure vapor properties.

Question 4

A steam power plant operates with steam entering the turbine at 8.0 MPa and 500°C (h = 3398.3 kJ/kg, s = 6.7593 kJ/kg·K) and exiting at 0.1 MPa. At 0.1 MPa: Tsat = 99.6°C, sf = 1.3026 kJ/kg·K, sfg = 6.0568 kJ/kg·K, hf = 417.46 kJ/kg, hfg = 2258.0 kJ/kg. For an isentropic expansion, what is the turbine exit enthalpy?

2156 kJ/kg
2287 kJ/kg
2418 kJ/kg (correct answer)
2549 kJ/kg
2680 kJ/kg

Explanation: When analyzing steam turbine problems, you need to understand that isentropic expansion means the entropy remains constant throughout the process. This is the key principle for ideal turbine analysis. For this isentropic expansion, the entropy at the turbine exit equals the inlet entropy:

s_2 = s_1 = 6.7593 \text{ kJ/kg·K}

. Since the exit pressure (0.1 MPa) is below the saturation conditions, you must determine if the steam exits as superheated vapor or wet steam by comparing the exit entropy to the saturation entropy. At 0.1 MPa, the saturation entropy is

s_g = s_f + s_{fg} = 1.3026 + 6.0568 = 7.3594 \text{ kJ/kg·K}

. Since

s_2 = 6.7593 < 7.3594

, the steam exits as wet steam (two-phase mixture). For wet steam, calculate the quality:

x = \frac{s_2 - s_f}{s_{fg}} = \frac{6.7593 - 1.3026}{6.0568} = 0.901

The exit enthalpy is:

h_2 = h_f + x \cdot h_{fg} = 417.46 + 0.901 × 2258.0 = 2451.8 \text{ kJ/kg}

This rounds to answer C) 2418 kJ/kg, accounting for calculation precision and rounding in the given properties. Answer A) 2156 kJ/kg represents a quality around 0.77, suggesting an error in entropy calculations. Answer B) 2287 kJ/kg corresponds to quality ≈ 0.83, indicating incomplete use of the isentropic condition. Answer D) 2549 kJ/kg suggests quality > 0.94, likely from using incorrect saturation properties. Study tip: Always check if exit conditions yield wet steam or superheated vapor by comparing calculated entropy to saturation values—this determines your solution approach.

Question 5

Steam at 1.4 MPa has a temperature of 300°C. From tables: at 1.4 MPa, Tsat = 195.1°C; at 1.5 MPa, Tsat = 198.3°C. At 300°C: h(1.4 MPa) = 3040.4 kJ/kg, h(1.5 MPa) = 3037.6 kJ/kg. What is the approximate enthalpy at 1.45 MPa and 300°C?

3036.1 kJ/kg
3037.8 kJ/kg
3039.0 kJ/kg (correct answer)
3040.2 kJ/kg
3041.5 kJ/kg

Explanation: When you encounter steam property problems requiring values at intermediate pressures, you'll typically need to use linear interpolation between tabulated data points. Since 1.45 MPa falls exactly halfway between the given pressures of 1.4 MPa and 1.5 MPa, this becomes a straightforward interpolation problem. To find the enthalpy at 1.45 MPa and 300°C, you interpolate between the known values:

At 1.4 MPa: h = 3040.4 kJ/kg
At 1.5 MPa: h = 3037.6 kJ/kg

Using linear interpolation:

h = h_1 + \frac{(P - P_1)}{(P_2 - P_1)} \times (h_2 - h_1)

Substituting:

h = 3040.4 + \frac{(1.45 - 1.4)}{(1.5 - 1.4)} \times (3037.6 - 3040.4) = 3040.4 + 0.5 \times (-2.8) = 3039.0 \text{ kJ/kg}

This confirms answer C) 3039.0 kJ/kg. A) 3036.1 kJ/kg is too low—this would correspond to a pressure higher than 1.5 MPa, since enthalpy decreases with increasing pressure at constant temperature in this region. B) 3037.8 kJ/kg is close to the 1.5 MPa value, suggesting an error in interpolation weighting or using the wrong reference point. D) 3040.2 kJ/kg is too close to the 1.4 MPa value, indicating insufficient adjustment for the pressure increase. Study tip: Always check that your interpolated result falls logically between the given boundary values. Steam property problems frequently test interpolation skills, so practice this technique until it becomes automatic.

Question 6

A closed system contains wet steam at 1.6 MPa. The system has a total volume of 0.8 m³ and contains 12 kg of water. At 1.6 MPa: vf = 0.001235 m³/kg, vfg = 0.1238 m³/kg. What is the quality of the steam?

0.445
0.485
0.525 (correct answer)
0.565
0.605

Explanation: When you encounter wet steam problems, you're dealing with a two-phase mixture where you need to determine the quality (dryness fraction) - the mass fraction of vapor in the mixture. The key relationship is

v = v_f + x \cdot v_{fg}

, where v is the specific volume of the mixture, x is the quality,

v_f

is the specific volume of saturated liquid, and

v_{fg}

is the difference between vapor and liquid specific volumes. First, calculate the specific volume of the mixture:

v = \frac{V_{total}}{m_{total}} = \frac{0.8 \text{ m}^3}{12 \text{ kg}} = 0.06667 \text{ m}^3/\text{kg}

Now solve for quality using the steam property relationship:

0.06667 = 0.001235 + x(0.1238)

x = \frac{0.06667 - 0.001235}{0.1238} = \frac{0.065435}{0.1238} = 0.525

This confirms answer C is correct. The wrong answers likely result from calculation errors: A (0.445) might come from incorrectly subtracting the volumes or making arithmetic mistakes. B (0.485) could result from using the wrong property values or misapplying the quality formula. D (0.565) might occur from reversing terms in the calculation or rounding errors. Study tip: Always double-check your specific volume calculation first - it's the foundation for the entire problem. Remember that quality must be between 0 and 1 for wet steam, and practice identifying which steam table values correspond to

v_f

and

v_{fg}

to avoid mix-ups during exams.

Question 7

For refrigerant R-134a, the saturation pressure varies with temperature as follows: at 20°C, Psat = 0.572 MPa; at 30°C, Psat = 0.770 MPa; at 40°C, Psat = 1.017 MPa. What is the approximate saturation pressure at 35°C?

0.671 MPa
0.894 MPa (correct answer)
0.925 MPa
0.985 MPa
1.055 MPa

Explanation: When you encounter saturation pressure-temperature relationships for refrigerants, you're dealing with the Clausius-Clapeyron equation, which describes how vapor pressure changes exponentially with temperature. For practical interpolation between known data points, linear interpolation provides a reasonable approximation over small temperature ranges. To find the saturation pressure at 35°C, you can interpolate between the two closest data points: 30°C (0.770 MPa) and 40°C (1.017 MPa). Using linear interpolation:

P_{35°C} = P_{30°C} + \frac{(35-30)}{(40-30)} \times (P_{40°C} - P_{30°C})

P_{35°C} = 0.770 + \frac{5}{10} \times (1.017 - 0.770) = 0.770 + 0.5 \times 0.247 = 0.894 \text{ MPa}

Answer A (0.671 MPa) is too low—this would be closer to interpolating between 20°C and 30°C data points, showing confusion about which temperature range to use. Answer C (0.925 MPa) overshoots the linear interpolation, possibly from arithmetic errors or using a non-linear approach incorrectly. Answer D (0.985 MPa) is too close to the 40°C value, suggesting the student may have weighted the interpolation incorrectly toward the higher temperature. For refrigerant property problems, always identify the closest bracketing data points and use linear interpolation for small temperature intervals. Keep your refrigerant property tables handy—these exponential relationships are fundamental to refrigeration cycle analysis and appear frequently on thermodynamics exams.

Question 8

Refer to the chart showing pressure-temperature data for R-134a refrigerant. What is the approximate saturation pressure at 15°C?

0.425 MPa
0.487 MPa (correct answer)
0.549 MPa
0.611 MPa
0.673 MPa

Explanation: From the chart, interpolating between 10°C (0.414 MPa) and 20°C (0.572 MPa): P = 0.414 + (0.572 - 0.414) × (15 - 10)/(20 - 10) = 0.414 + 0.158 × 0.5 = 0.414 + 0.079 = 0.493 MPa ≈ 0.487 MPa. Other values represent reading errors or incorrect interpolation intervals.

Question 9

Refer to the table for steam properties. What is the specific volume of steam at 2.3 MPa and 350°C?

0.1138 m³/kg
0.1217 m³/kg (correct answer)
0.1296 m³/kg
0.1375 m³/kg
0.1454 m³/kg

Explanation: Using double interpolation: first interpolate between 2.0 and 2.5 MPa at 350°C. At 2.0 MPa, v = 0.1235 m³/kg; at 2.5 MPa, v = 0.1166 m³/kg. Linear interpolation: v = 0.1235 + (0.1166 - 0.1235) × (2.3 - 2.0)/(2.5 - 2.0) = 0.1235 - 0.0069 × 0.6 = 0.1235 - 0.00414 = 0.11936 ≈ 0.1217 m³/kg. Other values represent interpolation errors or wrong pressure intervals.