A Platinum Thermometer

Q. The resistance of the platinum wire of a platinum resistance thermometer at the ice point is 5 Ω and at steam point is 5.23 Ω. When the thermometer is inserted in a hot bath, the resistance of the platinum wire is 5.795 Ω. What is the temperature of the bath, in °C? expects 345.65 °C

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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?

The playground

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}

The solution

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 Ruby

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

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

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

The laws

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)