A Steel Tape on a Hot Day

solved ThermalExpansion

Q. A steel tape 1m long is correctly calibrated for a temperature of 27.0 °C $T_{1,t} = 27.0$ °C. The length of a steel rod measured by this tape $L_{2,r} = L_{2,t}$ is found to be 63.0 cm $L_{1,t} = 63.0$ cm on a hot day when the temperature is 45.0 °C $T_{2,t} = 45.0$ °C. Coefficient of linear expansion of steel = 1.20 × 10⁻⁵ K⁻¹ $α_{l,t} = 1.20 \times 10^{−5}$ K⁻¹. What is the actual length of the steel rod on that day, in cm?

Inputs

calibrated

T_{1,t}

27 °C

reading

L_{1,t}

63 cm

hot

T_{2,t}

45 °C

expansion

α_{l,t}

1.2e-05 K⁻¹

Outputs

actual

L_{2,r}

How it's solved

Solved with ThermalExpansion

Given

$T_{1,t} = 27.0$ °C
$L_{1,t} = 63.0$ cm
$T_{2,t} = 45.0$ °C
$α_{l,t} = 1.20 \times 10^{−5}$ K⁻¹
$L_{2,r} = L_{2,t}$ from “measured by this tape”

Find $L_{2,r}$ .

Working

$T_{1,t} = 27.0$ °C $= 300.15$ K
$L_{1,t} = 63.0$ cm $= 0.63$ m
$T_{2,t} = 45.0$ °C $= 318.15$ K
linear expansion(tape) $L_{2,t} - L_{1,t} = α_{l,t} \cdot L_{1,t} \cdot (T_{2,t} - T_{1,t})$
$L_{2,t} - 0.63 = 1.2 \times 10^{−5} \cdot 0.63 \cdot (318.15 - 300.15)$
$L_{2,t} = 0.63013608$ m
measured by this tape $L_{2,r} = L_{2,t}$
$L_{2,r} = 0.63013608$
$L_{2,r} = 63.01$ cm
$L_{2,r} = 0.63013608$ m $= 63.013608$ cm
$L_{2,r} \approx 63.0136$ cmto 6 significant figures

Conclusion

On the hot day the tape has grown, so each of its marked centimetres is a little longer than a true one; the rod is as long as that stretched piece of tape.

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_steel_tape_on_a_hot_day do4  kind :numerical5  describes "A steel tape 1m long is correctly calibrated for a temperature of %{calibrated}. The length of a steel " \6            "rod measured by this tape is found to be %{reading} on a hot day when the temperature is %{hot}. " \7            "Coefficient of linear expansion of steel = %{expansion}. %{actual}, in cm?"8  given calibrated: "27.0 °C"9  given reading: "63.0 cm"10  given hot: "45.0 °C"11  given expansion: "1.20e-5 K-1"12  asks actual: "What is the actual length of the steel rod on that day"13end

solution.rb

1require_relative "../questions/a_steel_tape_on_a_hot_day"2require_relative "../science/thermal_properties_of_matter/thermal_expansion"34solve :a_steel_tape_on_a_hot_day, using: ThermalExpansion do5  objects :tape, :rod67  take calibrated: tape.temperature_at_first8  take reading: tape.length_at_first9  take hot: tape.temperature_after10  take expansion: tape.coefficient_of_linear_expansion1112  deduce_from "measured by this tape", rod.length_after => tape.length_after1314  actual = rod.length_after15  answer actual:, in: "cm", figures: 61617  conclusion "On the hot day the tape has grown, so each of its marked centimetres is a little longer " \18             "than a true one; the rod is as long as that stretched piece of tape."19end

playground.rb

1require_relative "../solutions/a_steel_tape_on_a_hot_day"23play_with :a_steel_tape_on_a_hot_day do4  title "A Steel Tape on a Hot Day"5end

The laws it drew on

quantity	written	units	within	from
length at first	$L_{1}$	m	—	ThermalExpansion
length after	$L_{2}$	m	—	ThermalExpansion
change in length	$Δ l$	m	—	ThermalExpansion
temperature at first	$T_{1}$	K	—	ThermalExpansion
temperature after	$T_{2}$	K	—	ThermalExpansion
coefficient of linear expansion	$α_{l}$	K⁻¹	—	ThermalExpansion
length after of ground	$L_{2,g}$	m	—
length at first of ground	$L_{1,g}$	m	—
coefficient of linear expansion of ground	$α_{l,g}$	K⁻¹	—
temperature after of ground	$T_{2,g}$	K	—
temperature at first of ground	$T_{1,g}$	K	—
length after of tape	$L_{2,t}$	m	—
length at first of tape	$L_{1,t}$	m	—
coefficient of linear expansion of tape	$α_{l,t}$	K⁻¹	—
temperature after of tape	$T_{2,t}$	K	—
temperature at first of tape	$T_{1,t}$	K	—
length after of rod	$L_{2,r}$	m	—
length at first of rod	$L_{1,r}$	m	—
coefficient of linear expansion of rod	$α_{l,r}$	K⁻¹	—
temperature after of rod	$T_{2,r}$	K	—
temperature at first of rod	$T_{1,r}$	K	—
change in length of ground	$Δ l_{g}$	m	—
change in length of tape	$Δ l_{t}$	m	—
change in length of rod	$Δ l_{r}$	m	—

L_{2,b} - L_{1,b} = α_{l,b} \cdot L_{1,b} \cdot (T_{2,b} - T_{1,b})

linear expansion

Δ l_{b} = L_{2,b} - L_{1,b}

the change is the length after less the length at first

L_{2,g} - L_{1,g} = α_{l,g} \cdot L_{1,g} \cdot (T_{2,g} - T_{1,g})

linear expansion(ground)

L_{2,t} - L_{1,t} = α_{l,t} \cdot L_{1,t} \cdot (T_{2,t} - T_{1,t})

linear expansion(tape)

L_{2,r} - L_{1,r} = α_{l,r} \cdot L_{1,r} \cdot (T_{2,r} - T_{1,r})

linear expansion(rod)

Δ l_{g} = L_{2,g} - L_{1,g}

the change is the length after less the length at first(ground)

Δ l_{t} = L_{2,t} - L_{1,t}

the change is the length after less the length at first(tape)

Δ l_{r} = L_{2,r} - L_{1,r}

the change is the length after less the length at first(rod)

L_{2,r} = L_{2,t}

measured by this tape