A Platinum Thermometer

solved TemperatureDependenceOfResistance

Q. The resistance of the platinum wire $α_{b} = α_{s}$ , $R_{0,b} = R_{0,s}$ of a platinum resistance thermometer at the ice point $T_{0} = 0$ °C is 5 Ω $R_{0,s} = 5$ Ω and at steam point $T_{s} = 100$ °C is 5.23 Ω $R_{s} = 5.23$ Ω. When the thermometer is inserted in a hot bath, the resistance of the platinum wire is 5.795 Ω $R_{b} = 5.795$ Ω. What is the temperature of the bath, in °C?

Inputs

at ice

R_{0,s}

5 Ω

at steam

R_{s}

5.23 Ω

in bath

R_{b}

5.795 Ω

Outputs

temperature

T_{b}

How it's solved

Solved with TemperatureDependenceOfResistance

Given

$R_{0,s} = 5$ Ω
$R_{s} = 5.23$ Ω
$R_{b} = 5.795$ Ω
$T_{0} = 0$ °C from “the ice point”
$T_{s} = 100$ °C from “steam point”
$α_{b} = α_{s}$ from “the platinum wire”
$R_{0,b} = R_{0,s}$ from “the platinum wire”

Find $T_{b}$ .

Working

$T_{0} = 0$ °C $= 273.15$ K
$T_{s} = 100$ °C $= 373.15$ K
the platinum wire (reference resistance of bath) $R_{0,b} = R_{0,s}$
$R_{0,b} = 5$
$R_{0,b} = 5$ Ω
resistance grows with temperature(steam) $R_{s} = R_{0,s} \cdot (1 + α_{s} \cdot (T_{s} - T_{0}))$
$5.23 = 5 \cdot (1 + α_{s} \cdot (373.15 - 273.15))$
$α_{s} = 4.6 \times 10^{−4}$ K⁻¹
the platinum wire (temperature coefficient of bath) $α_{b} = α_{s}$
$α_{b} = 4.6 \times 10^{−4}$
$α_{b} = 4.6 \times 10^{−4}$ K⁻¹
resistance grows with temperature(bath) $R_{b} = R_{0,b} \cdot (1 + α_{b} \cdot (T_{b} - T_{0}))$
$5.795 = 5 \cdot (1 + 4.6 \times 10^{−4} \cdot (T_{b} - 273.15))$
$T_{b} = 345.7$ °C
$T_{b} = 618.802174$ K $= 345.652174$ °C
$T_{b} \approx 345.65$ °Cto 5 significant figures

Conclusion

Platinum's resistance grows evenly with temperature, so the ice and steam points mark out a scale, and the bath's resistance reads off where it falls on it.

The web of concepts

given read from the words equation asked

The Ruby that solves it

The question and its solution are small Ruby files. The engine runs them.

question.rb

1require_relative "../physics"23question :a_platinum_thermometer do4  kind :numerical5  describes "The resistance of the platinum wire of a platinum resistance thermometer at the ice point is " \6            "%{at_ice} and at steam point is %{at_steam}. When the thermometer is inserted in a hot bath, the " \7            "resistance of the platinum wire is %{in_bath}. %{temperature}, in °C?"8  given at_ice: "5 Ω"9  given at_steam: "5.23 Ω"10  given in_bath: "5.795 Ω"11  asks temperature: "What is the temperature of the bath"12end

solution.rb

1require_relative "../questions/a_platinum_thermometer"2require_relative "../science/current_electricity/temperature_dependence_of_resistance"34solve :a_platinum_thermometer, using: TemperatureDependenceOfResistance do5  objects :steam, :bath6  deduce_from "the platinum wire", bath.temperature_coefficient => steam.temperature_coefficient,7                                   bath.reference_resistance => steam.reference_resistance89  take at_ice: steam.reference_resistance10  take at_steam: steam.resistance11  take in_bath: bath.resistance1213  deduce_from "the ice point", T_0 => "0 °C"14  deduce_from "steam point", steam.temperature => "100 °C"1516  temperature = bath.temperature17  answer temperature:, in: "°C", figures: 51819  conclusion "Platinum's resistance grows evenly with temperature, so the ice and steam points mark out a " \20             "scale, and the bath's resistance reads off where it falls on it."21end

playground.rb

1require_relative "../solutions/a_platinum_thermometer"23play_with :a_platinum_thermometer do4  title "A Platinum Thermometer"5end

The laws it drew on

quantity	written	units	within	from
voltage	$V$	V	—	OhmsLaw
current	$I$	A	—	OhmsLaw
resistance	$R$	Ω	0 … ∞	OhmsLaw
length of the conductor	$l$	m	—	OhmsLaw
cross section	$A$	m²	—	OhmsLaw
resistivity	$ρ$	Ω m	—	OhmsLaw
temperature coefficient	$α$	K⁻¹	—	TemperatureDependenceOfResistance
reference resistance	$R_{0}$	Ω	—	TemperatureDependenceOfResistance
reference temperature	$T_{0}$	K	—	TemperatureDependenceOfResistance
temperature	$T$	K	0 … ∞	TemperatureDependenceOfResistance
voltage of ground	$V_{g}$	V	—
current of ground	$I_{g}$	A	—
resistance of ground	$R_{g}$	Ω	0 … ∞
voltage of steam	$V_{s}$	V	—
current of steam	$I_{s}$	A	—
resistance of steam	$R_{s}$	Ω	0 … ∞
voltage of bath	$V_{b}$	V	—
current of bath	$I_{b}$	A	—
resistance of bath	$R_{b}$	Ω	0 … ∞
resistivity of ground	$ρ_{g}$	Ω m	—
length of the conductor of ground	$l_{g}$	m	—
cross section of ground	$A_{g}$	m²	—
resistivity of steam	$ρ_{s}$	Ω m	—
length of the conductor of steam	$l_{s}$	m	—
cross section of steam	$A_{s}$	m²	—
resistivity of bath	$ρ_{b}$	Ω m	—
length of the conductor of bath	$l_{b}$	m	—
cross section of bath	$A_{b}$	m²	—
reference resistance of ground	$R_{0,g}$	Ω	—
temperature coefficient of ground	$α_{g}$	K⁻¹	—
temperature of ground	$T_{g}$	K	0 … ∞
reference resistance of steam	$R_{0,s}$	Ω	—
temperature coefficient of steam	$α_{s}$	K⁻¹	—
temperature of steam	$T_{s}$	K	0 … ∞
reference resistance of bath	$R_{0,b}$	Ω	—
temperature coefficient of bath	$α_{b}$	K⁻¹	—
temperature of bath	$T_{b}$	K	0 … ∞

V_{c} = I_{c} \cdot R_{c}

ohms law

R_{c} = \frac{ρ_{c} \cdot l_{c}}{A_{c}}

resistance from resistivity

R_{c} = R_{0,c} \cdot (1 + α_{c} \cdot (T_{c} - T_{0}))

resistance grows with temperature

V_{g} = I_{g} \cdot R_{g}

ohms law(ground)

V_{s} = I_{s} \cdot R_{s}

ohms law(steam)

V_{b} = I_{b} \cdot R_{b}

ohms law(bath)

R_{g} = \frac{ρ_{g} \cdot l_{g}}{A_{g}}

resistance from resistivity(ground)

R_{s} = \frac{ρ_{s} \cdot l_{s}}{A_{s}}

resistance from resistivity(steam)

R_{b} = \frac{ρ_{b} \cdot l_{b}}{A_{b}}

resistance from resistivity(bath)

R_{g} = R_{0,g} \cdot (1 + α_{g} \cdot (T_{g} - T_{0}))

resistance grows with temperature(ground)

R_{s} = R_{0,s} \cdot (1 + α_{s} \cdot (T_{s} - T_{0}))

resistance grows with temperature(steam)

R_{b} = R_{0,b} \cdot (1 + α_{b} \cdot (T_{b} - T_{0}))

resistance grows with temperature(bath)

α_{b} = α_{s}

the platinum wire (temperature coefficient of bath)

R_{0,b} = R_{0,s}

the platinum wire (reference resistance of bath)