Temperature Scales and Thermometric Properties

18 questions

Question 1Question

Match each thermometer type listed on the left with its corresponding thermometric property on the right.

Click a left item, then click its matching right item

Items

Liquid-in-glass thermometer
Constant-volume gas thermometer
Platinum resistance thermometer
Thermocouple

Matches

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Answer

Liquid-in-glass thermometer matches change in volume of a liquid column; Constant-volume gas thermometer matches change in gas pressure; Platinum resistance thermometer matches change in electrical resistance; Thermocouple matches change in electromotive force (e.m.f.).
Each thermometer relies on a physical property that changes linearly or predictably with temperature: liquid-in-glass uses volume expansion of liquid, constant-volume gas thermometer uses gas pressure variation, platinum resistance thermometer uses electrical resistance change, and thermocouple uses electromotive force generated across thermal junctions.

Step-by-Step Solution

1
Identify the thermometric property for a liquid-in-glass thermometer.
Expansion of liquid volume.
The liquid (mercury or alcohol) expands up a narrow capillary tube as temperature increases.
2
Identify the thermometric property for a constant-volume gas thermometer.
Pressure of a gas.
At constant volume, the pressure of an ideal gas changes linearly with absolute temperature.
3
Identify the thermometric property for a platinum resistance thermometer.
Electrical resistance.
The electrical resistance of metals increases predictably with temperature.
4
Identify the thermometric property for a thermocouple.
Electromotive force (e.m.f.).
A temperature difference between two thermoelectric junctions induces a proportional voltage.

Key Concept

Thermometric properties of common thermometers
Estimated Time:45s
Question 2Question

A resistance thermometer has a resistance of 4.0Ω4.0\,\Omega at the ice point (0C0^\circ\text{C}) and 6.0Ω6.0\,\Omega at the steam point (100C100^\circ\text{C}). When immersed in a liquid bath, its resistance is measured to be 5.2Ω5.2\,\Omega. What is the temperature of the bath?

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Answer: 60.0C60.0^\circ\text{C}

Answer

60.0C60.0^\circ\text{C}
The temperature of the bath is found by calculating the fraction of resistance change relative to the total change between the ice point and steam point: θ=5.24.06.04.0×100C=60.0C\theta = \frac{5.2 - 4.0}{6.0 - 4.0} \times 100^\circ\text{C} = 60.0^\circ\text{C}.

Step-by-Step Solution

1
Identify the given thermometric parameters
R0=4.0ΩR_0 = 4.0\,\Omega, R100=6.0ΩR_{100} = 6.0\,\Omega, and Rθ=5.2ΩR_\theta = 5.2\,\Omega
These are the measured resistance values corresponding to the lower fixed point, upper fixed point, and unknown temperature.
2
Apply the linear temperature interpolation formula for a resistance thermometer
θ=RθR0R100R0×100C\theta = \frac{R_\theta - R_0}{R_{100} - R_0} \times 100^\circ\text{C}
Temperature on the Celsius scale is proportional to the relative change in the thermometric property between fixed points.
3
Substitute the values and compute the temperature
θ=5.24.06.04.0×100C=1.22.0×100C=60.0C\theta = \frac{5.2 - 4.0}{6.0 - 4.0} \times 100^\circ\text{C} = \frac{1.2}{2.0} \times 100^\circ\text{C} = 60.0^\circ\text{C}
Evaluating the expression yields the exact temperature of the liquid bath.

Key Concept

Linear interpolation on temperature scales using thermometric properties
Question 3Question

The length of the mercury column in an uncalibrated liquid-in-glass thermometer is 2.0cm2.0\,\text{cm} at the ice point (0C0^\circ\text{C}) and 18.0cm18.0\,\text{cm} at the steam point (100C100^\circ\text{C}). What is the temperature when the length of the mercury column is 9.2cm9.2\,\text{cm}?

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Answer: 45C45^\circ\text{C}

Answer

The temperature corresponding to a mercury column length of 9.2cm9.2\,\text{cm} is 45C45^\circ\text{C}.
The temperature on the Celsius scale is given by the formula θ=LθL0L100L0×100C\theta = \frac{L_\theta - L_0}{L_{100} - L_0} \times 100^\circ\text{C}. Substituting L0=2.0cmL_0 = 2.0\,\text{cm}, L100=18.0cmL_{100} = 18.0\,\text{cm}, and Lθ=9.2cmL_\theta = 9.2\,\text{cm} yields θ=7.216.0×100C=45C\theta = \frac{7.2}{16.0} \times 100^\circ\text{C} = 45^\circ\text{C}.

Step-by-Step Solution

1
Identify the fixed points and the thermometric property values.
Ice point length L0=2.0cmL_0 = 2.0\,\text{cm}, steam point length L100=18.0cmL_{100} = 18.0\,\text{cm}, and observed length Lθ=9.2cmL_\theta = 9.2\,\text{cm}.
The linear scale equation requires establishing the reference points on the Celsius scale.
2
Calculate the fundamental interval length (L100L0L_{100} - L_0).
L100L0=18.0cm2.0cm=16.0cmL_{100} - L_0 = 18.0\,\text{cm} - 2.0\,\text{cm} = 16.0\,\text{cm}.
The fundamental interval represents the total change in length corresponding to 100C100^\circ\text{C}.
3
Apply the linear interpolation formula θ=LθL0L100L0×100C\theta = \frac{L_\theta - L_0}{L_{100} - L_0} \times 100^\circ\text{C}.
θ=9.22.016.0×100C=7.216.0×100C=45C\theta = \frac{9.2 - 2.0}{16.0} \times 100^\circ\text{C} = \frac{7.2}{16.0} \times 100^\circ\text{C} = 45^\circ\text{C}.
This scales the fractional change in thermometric property above the ice point to degrees Celsius.

Key Concept

Linear interpolation on empirical temperature scales
Estimated Time:1m 30s
Question 4Question

Match each experimental temperature measurement requirement on the left with the most appropriate thermometric instrument on the right based on its thermometric property and operational characteristics.

Click a left item, then click its matching right item

Items

Standard calibration reference requiring high accuracy over a wide range using pressure variations at constant volume
High-precision steady-state measurement using electrical resistance variation where slight thermal response lag is permissible
Measurement of rapidly changing temperatures at a localized point using thermal electromotive force (e.m.f.)
Non-contact measurement of extremely high temperatures of glowing bodies using radiant energy intensity

Matches

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Answer

The correct pairings match each measurement requirement to its corresponding thermometric instrument based on its fundamental thermometric property: standard reference calibration pairs with the constant-volume gas thermometer; high-precision steady measurement pairs with the platinum resistance thermometer; rapid localized temperature change measurement pairs with the thermocouple; and non-contact high-temperature measurement pairs with the optical pyrometer.
Each instrument is correctly matched according to the specific physical property that changes measurably with temperature (PP, RR, e.m.f., and radiation intensity) and its operational suitability.

Step-by-Step Solution

1
Analyze requirement 1: standard reference calibration using gas pressure at constant volume.
Identified thermometric property as pressure PP at constant volume VV, which defines the constant-volume gas thermometer.
Gas thermometers closely approximate the absolute thermodynamic scale and serve as calibration standards.
2
Analyze requirement 2: high-precision steady measurement using resistance variation with thermal lag.
Identified thermometric property as electrical resistance RR, which corresponds to the platinum resistance thermometer.
Platinum wire resistance changes predictably with temperature, providing high accuracy for stable temperatures.
3
Analyze requirement 3: rapid localized temperature measurement via thermal e.m.f.
Identified thermometric property as thermoelectric voltage (e.m.f.), which corresponds to the thermocouple.
The small thermal mass of thermocouple junctions allows low response times for fast transient measurements.
4
Analyze requirement 4: non-contact measurement of glowing bodies using radiation.
Identified physical principle as thermal radiation intensity, corresponding to the optical pyrometer.
Pyrometers detect infrared/visible radiation, avoiding structural melting associated with direct contact at extreme temperatures.

Key Concept

Thermometric Properties and Operational Limits of Thermometers
Estimated Time:2m 0s
Question 5Question

A thermometer calibrated on an arbitrary scale XX registers a lower fixed point of 10X-10^\circ\text{X} and an upper fixed point of 110X110^\circ\text{X}. What is the true temperature in degrees Celsius (C^\circ\text{C}) when this thermometer reads 20X20^\circ\text{X}?

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Answer: 25

Answer

The true temperature on the Celsius scale is 25C25^\circ\text{C}.
Using the relation XLFPXUFPXLFPX=θ100\frac{X - \text{LFP}_X}{\text{UFP}_X - \text{LFP}_X} = \frac{\theta}{100}, substituting X=20X = 20, LFPX=10\text{LFP}_X = -10, and UFPX=110\text{UFP}_X = 110 gives 20(10)110(10)=30120=0.25\frac{20 - (-10)}{110 - (-10)} = \frac{30}{120} = 0.25. Multiplying 0.250.25 by 100100 gives 25C25^\circ\text{C}.

Step-by-Step Solution

1
Set up the linear relationship between the arbitrary temperature scale XX and the Celsius scale
XLFPXUFPXLFPX=θLFPCUFPCLFPC\frac{X - \text{LFP}_X}{\text{UFP}_X - \text{LFP}_X} = \frac{\theta - \text{LFP}_C}{\text{UFP}_C - \text{LFP}_C}
Thermometric properties vary linearly with temperature between fixed points.
2
Substitute the given numerical values into the formula
20(10)110(10)=θ01000    30120=θ100\frac{20 - (-10)}{110 - (-10)} = \frac{\theta - 0}{100 - 0} \implies \frac{30}{120} = \frac{\theta}{100}
The lower fixed point on scale XX is 10X-10^\circ\text{X} and the upper fixed point is 110X110^\circ\text{X}.
3
Solve for the unknown temperature θ\theta in degrees Celsius
θ=30120×100=25C\theta = \frac{30}{120} \times 100 = 25^\circ\text{C}
Simplifying the fraction 30120\frac{30}{120} yields 14\frac{1}{4}, and 14×100=25\frac{1}{4} \times 100 = 25.

Key Concept

Linear interpolation and conversion between thermometric temperature scales
Question 6Question

A thermometer is calibrated on a custom scale, XX, where the ice point (0C0^\circ\text{C}) is marked as 10X-10^\circ\text{X} and the steam point (100C100^\circ\text{C}) is marked as 110X110^\circ\text{X}. What is the reading on this custom scale when a standard Celsius thermometer reads 35C35^\circ\text{C}?

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Answer: 32

Answer

32 °X
Using the linear temperature interpolation formula XLFPXUFPXLFPX=CLFPCUFPCLFPC\frac{X - \text{LFP}_X}{\text{UFP}_X - \text{LFP}_X} = \frac{C - \text{LFP}_C}{\text{UFP}_C - \text{LFP}_C}, substituting LFPX=10\text{LFP}_X = -10, UFPX=110\text{UFP}_X = 110, C=35C = 35, LFPC=0\text{LFP}_C = 0, and UFPC=100\text{UFP}_C = 100 gives X(10)110(10)=3501000\frac{X - (-10)}{110 - (-10)} = \frac{35 - 0}{100 - 0}. Simplifying gives X+10120=0.35\frac{X + 10}{120} = 0.35, leading to X+10=42X + 10 = 42, so X=32XX = 32^\circ\text{X}.

Step-by-Step Solution

1
Determine fundamental intervals for both temperature scales
Celsius fundamental interval = 1000=100C100 - 0 = 100^\circ\text{C}; Custom scale fundamental interval = 110(10)=120X110 - (-10) = 120^\circ\text{X}
Linear temperature scale interpolation requires calculating the total interval between the lower fixed point (LFP) and upper fixed point (UFP).
2
Set up the ratio equation between the two thermometric scales
X(10)120=350100\frac{X - (-10)}{120} = \frac{35 - 0}{100}
The fractional position of any given temperature relative to its fixed points must be equal on all linear scales.
3
Solve the algebraic equation for XX
X+10=120×0.35=42    X=32XX + 10 = 120 \times 0.35 = 42 \implies X = 32^\circ\text{X}
Isolating XX gives the corresponding reading on the custom temperature scale.

Key Concept

Linear Temperature Scale Conversion and Interpolation
Question 7Question

A constant-volume gas thermometer registers a pressure of 50kPa50\,\text{kPa} at the ice point (0C0^\circ\text{C}) and 70kPa70\,\text{kPa} at the steam point (100C100^\circ\text{C}). What is the temperature when the gas pressure measured by the thermometer is 62kPa62\,\text{kPa}?

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Answer: 60C60^\circ\text{C}

Answer

60C60^\circ\text{C}
The temperature θ\theta on the Celsius scale is given by the ratio of the change in thermometric property from the ice point to the total fundamental interval, scaled by 100. Substituting P0=50kPaP_0 = 50\,\text{kPa}, P100=70kPaP_{100} = 70\,\text{kPa}, and Pθ=62kPaP_\theta = 62\,\text{kPa} gives θ=62507050×100=1220×100=60C\theta = \frac{62 - 50}{70 - 50} \times 100 = \frac{12}{20} \times 100 = 60^\circ\text{C}.

Step-by-Step Solution

1
Identify the given thermometric property values at the fixed points and target state
P0=50kPaP_0 = 50\,\text{kPa}, P100=70kPaP_{100} = 70\,\text{kPa}, and Pθ=62kPaP_\theta = 62\,\text{kPa}
These represent the lower fixed point, upper fixed point, and unknown temperature reading respectively.
2
Apply the general thermometric scale conversion formula
θ=PθP0P100P0×100C\theta = \frac{P_\theta - P_0}{P_{100} - P_0} \times 100^\circ\text{C}
Temperature on the Celsius scale varies linearly with the thermometric property relative to fixed points.
3
Substitute the values into the formula and solve for θ\theta
θ=62507050×100=1220×100=60C\theta = \frac{62 - 50}{70 - 50} \times 100 = \frac{12}{20} \times 100 = 60^\circ\text{C}
Carrying out the arithmetic yields the exact temperature of 60C60^\circ\text{C}.

Key Concept

Linear interpolation on temperature scales using thermometric properties
Estimated Time:1m 0s
Question 8Question

A thermometric property XX of a system has a value of 15.0units15.0\,\text{units} at the ice point (0C0^\circ\text{C}) and 75.0units75.0\,\text{units} at the steam point (100C100^\circ\text{C}). If the thermometric property is measured as 33.0units33.0\,\text{units} when immersed in a chemical bath, what is the temperature of the bath on the absolute thermodynamic scale?

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Answer: 303K303\,\text{K}

Answer

303K303\,\text{K}
The temperature of the bath on the Celsius scale is found by taking the ratio of the change in thermometric property from the ice point to the total fundamental interval: 33.015.075.015.0×100C=1860×100=30C\frac{33.0 - 15.0}{75.0 - 15.0} \times 100^\circ\text{C} = \frac{18}{60} \times 100 = 30^\circ\text{C}. Converting this temperature to the absolute (Kelvin) scale requires adding 273K273\,\text{K}, yielding 30+273=303K30 + 273 = 303\,\text{K}.

Step-by-Step Solution

1
Identify the given thermometric values for the fixed points and the unknown state
X0=15.0unitsX_0 = 15.0\,\text{units} (ice point, 0C0^\circ\text{C}), X100=75.0unitsX_{100} = 75.0\,\text{units} (steam point, 100C100^\circ\text{C}), and XT=33.0unitsX_T = 33.0\,\text{units}
Linear thermometric interpolation requires defining the fixed reference points and the measured property value.
2
Calculate the temperature on the Celsius scale using the linear scale formula
θ=XTX0X100X0×100C=33.015.075.015.0×100=18.060.0×100=30C\theta = \frac{X_T - X_0}{X_{100} - X_0} \times 100^\circ\text{C} = \frac{33.0 - 15.0}{75.0 - 15.0} \times 100 = \frac{18.0}{60.0} \times 100 = 30^\circ\text{C}
The temperature change relative to the fundamental interval determines the position on the Celsius scale.
3
Convert the temperature from degrees Celsius to Kelvin
T=θ+273=30+273=303KT = \theta + 273 = 30 + 273 = 303\,\text{K}
The absolute thermodynamic scale (Kelvin) is shifted from the Celsius scale by adding 273K273\,\text{K}.

Key Concept

Linear interpolation of temperature using thermometric properties and conversion to the absolute scale
Estimated Time:2m 0s
Question 9Question

Match each type of thermometer with its corresponding physical thermometric property.

Click a left item, then click its matching right item

Items

Liquid-in-glass thermometer
Constant-volume gas thermometer
Resistance thermometer
Thermocouple

Matches

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Answer

Liquid-in-glass thermometer matches change in length or volume of a liquid column; Constant-volume gas thermometer matches change in pressure of a gas; Resistance thermometer matches change in electrical resistance; Thermocouple matches electromotive force (e.m.f.) produced across junctions.
Each thermometer is accurately matched to the physical property that undergoes a measurable change as temperature varies.

Step-by-Step Solution

1
Identify the defining physical property that varies with temperature for each instrument.
Each thermometer operates on a distinct physical property that changes predictably when heated or cooled.
Thermometric properties must be reproducible and continuously measurable across a temperature range.
2
Pair each instrument with its specific thermometric property.
Liquid-in-glass pairs with liquid column expansion/length; constant-volume gas thermometer pairs with gas pressure; resistance thermometer pairs with electrical resistance; thermocouple pairs with thermoelectric e.m.f.
These pairs represent standard physical principles used in thermometry.

Key Concept

Thermometric properties and operating principles of thermometers
Question 10Question

Match each temperature scale reference state on the left with its correct thermodynamic definition on the right.

Click a left item, then click its matching right item

Items

Absolute zero (0 K0\text{ K})
Ice point (0C0^\circ\text{C})
Steam point (100C100^\circ\text{C})
Triple point of water (273.16 K273.16\text{ K})

Matches

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Answer

Absolute zero matches with the state of minimum molecular kinetic energy. The ice point matches with the lower fixed point of pure melting ice at standard pressure. The steam point matches with the upper fixed point of pure boiling water steam at standard pressure. The triple point of water matches with the thermodynamic equilibrium state of ice, liquid water, and water vapour.
Each temperature scale reference point is correctly paired with its defining physical state: absolute zero represents minimum molecular kinetic energy, the ice point represents melting ice at standard pressure, the steam point represents steam from boiling water at standard pressure, and the triple point represents the three-phase equilibrium of water.

Step-by-Step Solution

1
Identify the definition of absolute zero.
Absolute zero (0 K0\text{ K}) corresponds to the state of minimum internal molecular kinetic energy.
At 0 K0\text{ K}, thermal motion of particles theoretically ceases.
2
Identify the definition of the ice point.
The ice point (0C0^\circ\text{C}) corresponds to pure melting ice at standard atmospheric pressure.
It serves as the standard lower fixed point on the Celsius temperature scale.
3
Identify the definition of the steam point.
The steam point (100C100^\circ\text{C}) corresponds to steam from pure boiling water at standard atmospheric pressure.
It serves as the standard upper fixed point on the Celsius temperature scale.
4
Identify the definition of the triple point of water.
The triple point (273.16 K273.16\text{ K}) is the unique thermodynamic state where ice, liquid water, and steam coexist in equilibrium.
It is used as a single fundamental reference point on the Kelvin thermodynamic scale.

Key Concept

Temperature Scale Fixed Points and Reference States
Estimated Time:1m 30s
Question 11Question

A resistance thermometer registers a resistance of 5.0Ω5.0\,\Omega at the ice point (0C0^\circ\text{C}) and 25.0Ω25.0\,\Omega at the steam point (100C100^\circ\text{C}). When placed in a heated liquid bath, the resistance measured is 30.0Ω30.0\,\Omega. What is the temperature of the bath on the Kelvin scale?

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Answer: 398K398\,\text{K}

Answer

The temperature of the bath on the Kelvin scale is 398K398\,\text{K}.
Using the linear interpolation formula for a thermometric property θ=RθR0R100R0×100C\theta = \frac{R_\theta - R_0}{R_{100} - R_0} \times 100^\circ\text{C}, substituting R0=5.0ΩR_0 = 5.0\,\Omega, R100=25.0ΩR_{100} = 25.0\,\Omega, and Rθ=30.0ΩR_\theta = 30.0\,\Omega yields θ=25.020.0×100=125C\theta = \frac{25.0}{20.0} \times 100 = 125^\circ\text{C}. Converting to absolute thermodynamic temperature gives T=125+273=398KT = 125 + 273 = 398\,\text{K}.

Step-by-Step Solution

1
Calculate the temperature on the Celsius scale using linear interpolation of thermometric property.
θ=RθR0R100R0×100C=30.05.025.05.0×100C=25.020.0×100C=125C\theta = \frac{R_\theta - R_0}{R_{100} - R_0} \times 100^\circ\text{C} = \frac{30.0 - 5.0}{25.0 - 5.0} \times 100^\circ\text{C} = \frac{25.0}{20.0} \times 100^\circ\text{C} = 125^\circ\text{C}
The change in resistance is directly proportional to the temperature change between fixed points.
2
Convert the temperature from degrees Celsius to Kelvins.
T=θ+273=125+273=398KT = \theta + 273 = 125 + 273 = 398\,\text{K}
Absolute temperature in Kelvin is obtained by adding 273 to the temperature in degrees Celsius.

Key Concept

Temperature Scale Interpolation and Kelvin Conversion
Estimated Time:1m 30s
Question 12Question

The length of the mercury column in an uncalibrated thermometer is 4.0cm4.0\,\text{cm} at the ice point (0C0^\circ\text{C}) and 24.0cm24.0\,\text{cm} at the steam point (100C100^\circ\text{C}). What is the temperature in degrees Celsius when the length of the mercury column is 19.0cm19.0\,\text{cm}?

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Answer: 75

Answer

The temperature corresponding to a mercury column length of 19.0cm19.0\,\text{cm} is 75C75^\circ\text{C}.
The temperature on the Celsius scale is determined by the ratio of the length change above the ice point to the total length change between the ice and steam points: T=LTL0L100L0×100CT = \frac{L_T - L_0}{L_{100} - L_0} \times 100^\circ\text{C}. Substituting L0=4.0cmL_0 = 4.0\,\text{cm}, L100=24.0cmL_{100} = 24.0\,\text{cm}, and LT=19.0cmL_T = 19.0\,\text{cm} gives T=15.020.0×100=75CT = \frac{15.0}{20.0} \times 100 = 75^\circ\text{C}.

Step-by-Step Solution

1
Identify given thermometric length values at fixed points and at the unknown temperature
L0=4.0cmL_0 = 4.0\,\text{cm}, L100=24.0cmL_{100} = 24.0\,\text{cm}, and LT=19.0cmL_T = 19.0\,\text{cm}
These represent the length at the lower fixed point (0C0^\circ\text{C}), upper fixed point (100C100^\circ\text{C}), and intermediate temperature TT respectively.
2
Set up the linear interpolation equation on the Celsius scale
T=LTL0L100L0×100CT = \frac{L_T - L_0}{L_{100} - L_0} \times 100^\circ\text{C}
Thermometric expansion is assumed to vary linearly with temperature over the operational range.
3
Substitute the given values and perform arithmetic calculation
T=19.04.024.04.0×100=15.020.0×100=75CT = \frac{19.0 - 4.0}{24.0 - 4.0} \times 100 = \frac{15.0}{20.0} \times 100 = 75^\circ\text{C}
Simplifying 15.020.0\frac{15.0}{20.0} gives 0.750.75, which multiplied by 100100 equals 7575.

Key Concept

Temperature measurement using linear variation of thermometric properties
Question 13Question

A faulty liquid-in-glass thermometer registers a reading of 5C5^\circ\text{C} at the melting ice point and 95C95^\circ\text{C} at the steam point of pure water under standard atmospheric pressure. What is the actual temperature in degrees Celsius when this thermometer registers a reading of 41C41^\circ\text{C}?

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Answer: 40C40^\circ\text{C}

Answer

The actual temperature is 40C40^\circ\text{C}.
The correct answer is obtained by setting up the linear interpolation formula for thermometric property values: θ=XθX0X100X0×100\theta = \frac{X_\theta - X_0}{X_{100} - X_0} \times 100. Substituting X0=5X_0 = 5, X100=95X_{100} = 95, and Xθ=41X_\theta = 41 gives θ=3690×100=40C\theta = \frac{36}{90} \times 100 = 40^\circ\text{C}.

Step-by-Step Solution

1
Determine the fundamental interval of the faulty thermometer.
Fundamental interval =95C5C=90divisions= 95^\circ\text{C} - 5^\circ\text{C} = 90\,\text{divisions}.
The total interval between the lower fixed point and upper fixed point represents 100C100^\circ\text{C} on the standard Celsius scale.
2
Calculate the measured change from the ice point.
Measured difference =41C5C=36divisions= 41^\circ\text{C} - 5^\circ\text{C} = 36\,\text{divisions}.
The zero error of +5C+5^\circ\text{C} must be subtracted from the observed reading.
3
Apply the linear scale interpolation formula to find the actual temperature θ\theta.
θ=415955×100C=3690×100C=40C\theta = \frac{41 - 5}{95 - 5} \times 100^\circ\text{C} = \frac{36}{90} \times 100^\circ\text{C} = 40^\circ\text{C}.
The ratio of the measured interval to the total fundamental interval equals the true fraction of 100C100^\circ\text{C}.

Key Concept

Linear interpolation on non-standard or faulty thermometer scales using fixed points.
Question 14Question

A platinum resistance thermometer has a resistance of 10Ω10\,\Omega at the ice point (0C0^\circ\text{C}) and 50Ω50\,\Omega at the steam point (100C100^\circ\text{C}). When placed in a warm liquid bath, an uncalibrated digital ohmmeter reads 38Ω38\,\Omega. If the ohmmeter has a known positive zero error of +4Ω+4\,\Omega (indicating 4Ω4\,\Omega above the true resistance), what is the actual temperature of the liquid bath on the absolute scale in kelvins?

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Answer: 333K333\,\text{K}

Answer

The actual temperature of the liquid bath on the absolute scale is 333K333\,\text{K}.
Subtracting the zero error of +4Ω+4\,\Omega from the raw meter reading of 38Ω38\,\Omega yields the true thermometric resistance of 34Ω34\,\Omega. Applying the temperature formula θ=RθR0R100R0×100C\theta = \frac{R_\theta - R_0}{R_{100} - R_0} \times 100^\circ\text{C} gives θ=34105010×100=60C\theta = \frac{34 - 10}{50 - 10} \times 100 = 60^\circ\text{C}. Converting to absolute temperature gives T=60+273=333KT = 60 + 273 = 333\,\text{K}.

Step-by-Step Solution

1
Correct the measured resistance for zero error
True resistance Rθ=38Ω4Ω=34ΩR_\theta = 38\,\Omega - 4\,\Omega = 34\,\Omega
A positive zero error means the meter reads higher than the true value, so the zero error must be subtracted.
2
Calculate the temperature on the Celsius scale using linear interpolation
θ=RθR0R100R0×100C=34105010×100=2440×100=60C\theta = \frac{R_\theta - R_0}{R_{100} - R_0} \times 100^\circ\text{C} = \frac{34 - 10}{50 - 10} \times 100 = \frac{24}{40} \times 100 = 60^\circ\text{C}
The thermometric property varies linearly between the fixed points.
3
Convert the Celsius temperature to the thermodynamic (absolute) Kelvin scale
T=θ+273=60+273=333KT = \theta + 273 = 60 + 273 = 333\,\text{K}
The conversion from Celsius to Kelvin requires adding 273273 (or 273.15273.15).

Key Concept

Linear interpolation of thermometric properties with instrument zero error correction
Estimated Time:2m 0s
Question 15Question

Match each thermometer type on the left with its corresponding physical thermometric property on the right.

Click a left item, then click its matching right item

Items

Constant-volume gas thermometer
Thermocouple
Platinum resistance thermometer
Liquid-in-glass thermometer

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Answer

Constant-volume gas thermometer matches variation of gas pressure at constant volume; Thermocouple matches variation of electromotive force (e.m.f.); Platinum resistance thermometer matches variation of electrical resistance; Liquid-in-glass thermometer matches variation of liquid column length/volume.
Each thermometer is paired with its corresponding thermometric property: constant-volume gas thermometer relies on gas pressure variation, thermocouple relies on thermoelectric e.m.f. variation, resistance thermometer relies on electrical resistance variation, and liquid-in-glass thermometer relies on liquid thermal expansion.

Step-by-Step Solution

1
Identify the thermometric property for a constant-volume gas thermometer.
Gas pressure at constant volume.
Pressure varies linearly with temperature for an ideal gas at constant volume.
2
Identify the thermometric property for a thermocouple.
Electromotive force (e.m.f.).
The thermoelectric effect generates an e.m.f. proportional to the temperature difference between two junctions.
3
Identify the thermometric property for a platinum resistance thermometer.
Electrical resistance.
The electrical resistance of platinum increases continuously and predictably with temperature.
4
Identify the thermometric property for a liquid-in-glass thermometer.
Liquid column length/volume.
The liquid expands linearly along the capillary stem as temperature rises.

Key Concept

Thermometric Properties of Thermometers
Question 16Question

A thermocouple thermometer produces an electromotive force (e.m.f.) of 2.0mV2.0\,\text{mV} at the ice point (0C0^\circ\text{C}) and 18.0mV18.0\,\text{mV} at the steam point (100C100^\circ\text{C}). When placed in a liquid bath, the recorded e.m.f. is 14.0mV14.0\,\text{mV}. What is the temperature of the liquid bath on the Celsius scale?

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Answer: 75.0C75.0^\circ\text{C}

Answer

75.0C75.0^\circ\text{C}
The temperature on a linear scale is proportional to the fraction of the interval traversed between the fixed points. Subtracting the baseline reading of 2.0mV2.0\,\text{mV} gives an effective increase of 12.0mV12.0\,\text{mV} out of a total range of 16.0mV16.0\,\text{mV}. Multiplying this fraction (0.750.75) by 100C100^\circ\text{C} yields 75.0C75.0^\circ\text{C}.

Step-by-Step Solution

1
Identify the given thermometric values for the lower fixed point, upper fixed point, and unknown temperature reading.
E0=2.0mVE_0 = 2.0\,\text{mV}, E100=18.0mVE_{100} = 18.0\,\text{mV}, and Eθ=14.0mVE_\theta = 14.0\,\text{mV}.
Linear temperature scales relate the change in thermometric property proportionally to temperature changes.
2
Apply the standard linear interpolation formula for a Celsius temperature scale.
θ=EθE0E100E0×100C\theta = \frac{E_\theta - E_0}{E_{100} - E_0} \times 100^\circ\text{C}
This accounts for the baseline reading at the ice point (0C0^\circ\text{C}) and normalizes it over the 100C100^\circ\text{C} fundamental interval.
3
Substitute the known values and evaluate the expression.
θ=14.02.018.02.0×100=12.016.0×100=0.75×100=75.0C\theta = \frac{14.0 - 2.0}{18.0 - 2.0} \times 100 = \frac{12.0}{16.0} \times 100 = 0.75 \times 100 = 75.0^\circ\text{C}.
Carrying out the subtraction yields a proportional change of 34\frac{3}{4} of the full fundamental interval.

Key Concept

Temperature Scale Calibration and Thermocouple Interpolation
Estimated Time:1m 30s
Question 17Question

A constant-volume gas thermometer registers a pressure of 60kPa60\,\text{kPa} at the ice point (0C0^\circ\text{C}) and 84kPa84\,\text{kPa} at the steam point (100C100^\circ\text{C}). What is the temperature in degrees Celsius when the pressure registered by the thermometer is 72kPa72\,\text{kPa}?

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Answer: 50

Answer

The temperature corresponding to a pressure reading of 72kPa72\,\text{kPa} is 50C50^\circ\text{C}.
Using the linear relation for a constant-volume gas thermometer: T=PTP0P100P0×100CT = \frac{P_T - P_0}{P_{100} - P_0} \times 100^\circ\text{C}. Substituting PT=72kPaP_T = 72\,\text{kPa}, P0=60kPaP_0 = 60\,\text{kPa}, and P100=84kPaP_{100} = 84\,\text{kPa} yields T=72608460×100=1224×100=50CT = \frac{72 - 60}{84 - 60} \times 100 = \frac{12}{24} \times 100 = 50^\circ\text{C}.

Step-by-Step Solution

1
Identify the thermometric property values at the fixed points and target state.
P0=60kPaP_0 = 60\,\text{kPa}, P100=84kPaP_{100} = 84\,\text{kPa}, and PT=72kPaP_T = 72\,\text{kPa}.
These represent the pressure values corresponding to 0C0^\circ\text{C}, 100C100^\circ\text{C}, and the unknown temperature TT respectively.
2
Set up the linear scale conversion equation.
T=PTP0P100P0×100CT = \frac{P_T - P_0}{P_{100} - P_0} \times 100^\circ\text{C}
Temperature changes linearly with the thermometric property (gas pressure at constant volume).
3
Calculate the numerical value.
T=1224×100=50CT = \frac{12}{24} \times 100 = 50^\circ\text{C}
Simplifying the fraction 1224=0.5\frac{12}{24} = 0.5 and multiplying by 100100 gives 5050.

Key Concept

Temperature measurement using constant-volume gas pressure as a thermometric property
Question 18Question

A thermistor has an electrical resistance of 800Ω800\,\Omega at the melting point of ice (0C0^\circ\text{C}) and 200Ω200\,\Omega at the boiling point of water (100C100^\circ\text{C}). Assuming the thermometric property varies linearly with temperature, what is the temperature in degrees Celsius (C^\circ\text{C}) when the resistance of the thermistor is 500Ω500\,\Omega?

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Answer: 50

Answer

The temperature corresponding to a resistance of 500Ω500\,\Omega is 50C50^\circ\text{C}.
Applying the standard thermometric interpolation relation θ=RθR0R100R0×100C\theta = \frac{R_\theta - R_0}{R_{100} - R_0} \times 100^\circ\text{C} with R0=800ΩR_0 = 800\,\Omega, R100=200ΩR_{100} = 200\,\Omega, and Rθ=500ΩR_\theta = 500\,\Omega gives θ=500800200800×100=300600×100=50C\theta = \frac{500 - 800}{200 - 800} \times 100 = \frac{-300}{-600} \times 100 = 50^\circ\text{C}.

Step-by-Step Solution

1
Identify the values of the thermometric property at the ice point and steam point.
R0=800ΩR_0 = 800\,\Omega and R100=200ΩR_{100} = 200\,\Omega.
These established values represent the fixed points of 0C0^\circ\text{C} and 100C100^\circ\text{C} respectively.
2
Set up the linear relationship formula for temperature conversion on the Celsius scale.
\(\theta = \frac{R_\theta - R_0}{R_{100} - R_0} \times 100^\circ\text{C}\)
The temperature scale is defined linearly between the two fixed calibration points.
3
Substitute the unknown resistance Rθ=500ΩR_\theta = 500\,\Omega into the equation and compute the result.
\(\theta = \frac{500 - 800}{200 - 800} \times 100^\circ\text{C} = \frac{-300}{-600} \times 100^\circ\text{C} = 50^\circ\text{C}\)
Dividing the change from the lower fixed point by the total interval between fixed points gives the fraction of 100C100^\circ\text{C}.

Key Concept

Linear relationship between a thermometric property and temperature
Temperature Scales and Thermometric Properties Practice Questions — JAMB UTME | Examkin