A Steel Tape on a Hot Day

Q. A steel tape 1m long is correctly calibrated for a temperature of 27.0 °C. The length of a steel rod measured by this tape is found to be 63.0 cm on a hot day when the temperature is 45.0 °C. Coefficient of linear expansion of steel = 1.20 × 10⁻⁵ K⁻¹. What is the actual length of the steel rod on that day, in cm? expects 63.0136 cm

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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?

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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}

The solution

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 Ruby

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

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

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

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