A Heating Element at 117 Ohms

Q. At room temperature (27.0 °C) the resistance of a heating element is 100 Ω. What is the temperature of the element if the resistance is found to be 117 Ω, given that the temperature coefficient of the material of the resistor is 1.70 × 10⁻⁴ °C⁻¹, in °C? expects 1027 °C

What the draft establishes

Fills in once a draft is checked.

The web of concepts

given read from the words equation asked

The equations it used

None yet.

Check an answer

Q. At room temperature (27.0 °C $T_{0} = 27.0$ °C) the resistance of a heating element is 100 Ω $R_{0} = 100$ Ω. What is the temperature of the element if the resistance is found to be 117 Ω $R = 117$ Ω, given that the temperature coefficient of the material of the resistor is 1.70 × 10⁻⁴ °C⁻¹ $α = 1.70 \times 10^{−4}$ °C⁻¹, in °C?

The playground

Inputs

room

T_{0}

27 °C

at room

R_{0}

100 Ω

hot

R

117 Ω

coefficient

α

0.00017 °C⁻¹

Outputs

temperature

T

The solution

Solved with TemperatureDependenceOfResistance

Given

$T_{0} = 27.0$ °C
$R_{0} = 100$ Ω
$R = 117$ Ω
$α = 1.70 \times 10^{−4}$ °C⁻¹

Find $T$ .

Working

$T_{0} = 27.0$ °C $= 300.15$ K
$α = 1.70 \times 10^{−4}$ °C⁻¹ $= 1.7 \times 10^{−4}$ K⁻¹
resistance grows with temperature(element) $R = R_{0} \cdot (1 + α \cdot (T - T_{0}))$
$117 = 100 \cdot (1 + 1.7 \times 10^{−4} \cdot (T - 300.15))$
$T = 1027$ °C
$T = 1300.15$ K $= 1027$ °C
$T \approx 1027$ °Cto 4 significant figures

Conclusion

The resistance grows in step with the rise in temperature, so how far it has grown tells how far the element is above room temperature.

The Ruby

1require_relative "../physics"23question :a_heating_element_at_117_ohms do4  kind :numerical5  describes "At room temperature (%{room}) the resistance of a heating element is %{at_room}. %{temperature} if the " \6            "resistance is found to be %{hot}, given that the temperature coefficient of the material of the " \7            "resistor is %{coefficient}, in °C?"8  given room: "27.0 °C"9  given at_room: "100 Ω"10  given hot: "117 Ω"11  given coefficient: "1.70e-4 °C-1"12  asks temperature: "What is the temperature of the element"13end

1require_relative "../questions/a_heating_element_at_117_ohms"2require_relative "../science/current_electricity/temperature_dependence_of_resistance"34solve :a_heating_element_at_117_ohms, using: TemperatureDependenceOfResistance do5  objects :element67  take room: T_08  take at_room: element.reference_resistance9  take hot: element.resistance10  take coefficient: element.temperature_coefficient1112  temperature = element.temperature13  answer temperature:, in: "°C", figures: 41415  conclusion "The resistance grows in step with the rise in temperature, so how far it has grown tells how far " \16             "the element is above room temperature."17end

1require_relative "../solutions/a_heating_element_at_117_ohms"23play_with :a_heating_element_at_117_ohms do4  title "A Heating Element at 117 Ohms"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 element	$V$	V	—
current of element	$I$	A	—
resistance of element	$R$	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 element	$ρ$	Ω m	—
length of the conductor of element	$l$	m	—
cross section of element	$A$	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 element	$R_{0}$	Ω	—
temperature coefficient of element	$α$	K⁻¹	—
temperature of element	$T$	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 = I \cdot R

ohms law(element)

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

resistance from resistivity(ground)

R = \frac{ρ \cdot l}{A}

resistance from resistivity(element)

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

resistance grows with temperature(ground)

R = R_{0} \cdot (1 + α \cdot (T - T_{0}))

resistance grows with temperature(element)