Iron and Brass End to End

solved Conduction

Q. An iron bar (length 0.1 m $L_{i} = 0.1$ m, cross-section 0.02 m² $A_{i} = 0.02$ m², thermal conductivity 79 W m⁻¹ K⁻¹ $K_{i} = 79$ W m⁻¹ K⁻¹) and a brass bar (length 0.1 m $L_{b} = 0.1$ m, cross-section 0.02 m² $A_{b} = 0.02$ m², thermal conductivity 109 W m⁻¹ K⁻¹ $K_{b} = 109$ W m⁻¹ K⁻¹) are soldered end to end $T_{C,b} = T_{D,i}$ , $H_{b} = H_{i}$ . The free ends of the iron bar and brass bar are maintained at 373 K $T_{C,i} = 373$ K and 273 K $T_{D,b} = 273$ K respectively. What is the temperature of the junction of the two bars, in kelvin?

Inputs

iron length

L_{i}

0.10 m

iron area

A_{i}

0.02 m²

iron k

K_{i}

79 W m⁻¹ K⁻¹

brass length

L_{b}

0.10 m

brass area

A_{b}

0.02 m²

brass k

K_{b}

109 W m⁻¹ K⁻¹

hot

T_{C,i}

373 K

cold

T_{D,b}

273 K

Outputs

junction

T_{D,i}

How it's solved

Solved with Conduction

Given

$L_{i} = 0.1$ m
$A_{i} = 0.02$ m²
$K_{i} = 79$ W m⁻¹ K⁻¹
$L_{b} = 0.1$ m
$A_{b} = 0.02$ m²
$K_{b} = 109$ W m⁻¹ K⁻¹
$T_{C,i} = 373$ K
$T_{D,b} = 273$ K
$T_{C,b} = T_{D,i}$ from “soldered end to end”
$H_{b} = H_{i}$ from “soldered end to end”

Find $T_{D,i}$ .

Working

heat flows down a rod(brass) $H_{b} = \frac{K_{b} \cdot A_{b} \cdot (T_{C,b} - T_{D,b})}{L_{b}}$
soldered end to end (heat current of brass) $H_{b} = H_{i}$
heat flows down a rod(iron) $H_{i} = \frac{K_{i} \cdot A_{i} \cdot (T_{C,i} - T_{D,i})}{L_{i}}$
soldered end to end (hot end of brass) $T_{C,b} = T_{D,i}$
$T_{D,i} = 315$ K
$T_{D,i} \approx 315$ Kto 3 significant figures

Conclusion

In the steady state the same heat flows through both bars, so the junction settles where the better conductor needs a smaller drop to carry 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 :iron_and_brass_end_to_end do4  kind :numerical5  describes "An iron bar (length %{iron_length}, cross-section %{iron_area}, thermal conductivity %{iron_k}) and a " \6            "brass bar (length %{brass_length}, cross-section %{brass_area}, thermal conductivity %{brass_k}) are " \7            "soldered end to end. The free ends of the iron bar and brass bar are maintained at %{hot} and %{cold} " \8            "respectively. %{junction}, in kelvin?"9  given iron_length: "0.1 m"10  given iron_area: "0.02 m2"11  given iron_k: "79 W m-1 K-1"12  given brass_length: "0.1 m"13  given brass_area: "0.02 m2"14  given brass_k: "109 W m-1 K-1"15  given hot: "373 K"16  given cold: "273 K"17  asks junction: "What is the temperature of the junction of the two bars"18end

solution.rb

1require_relative "../questions/iron_and_brass_end_to_end"2require_relative "../science/thermal_properties_of_matter/conduction"34solve :iron_and_brass_end_to_end, using: Conduction do5  objects :iron, :brass67  take iron_length: iron.length_of_the_rod8  take iron_area: iron.cross_section9  take iron_k: iron.thermal_conductivity10  take brass_length: brass.length_of_the_rod11  take brass_area: brass.cross_section12  take brass_k: brass.thermal_conductivity13  take hot: iron.hot_end14  take cold: brass.cold_end1516  deduce_from "soldered end to end", brass.hot_end => iron.cold_end, brass.heat_current => iron.heat_current1718  junction = iron.cold_end19  answer junction:, in: "K", figures: 32021  conclusion "In the steady state the same heat flows through both bars, so the junction settles where " \22             "the better conductor needs a smaller drop to carry it."23end

playground.rb

1require_relative "../solutions/iron_and_brass_end_to_end"23play_with :iron_and_brass_end_to_end do4  title "Iron and Brass End to End"5end

The laws it drew on

quantity	written	units	within	from
thermal conductivity	$K$	W m⁻¹ K⁻¹	—	Conduction
cross section	$A$	m²	—	Conduction
length of the rod	$L$	m	—	Conduction
hot end	$T_{C}$	K	—	Conduction
cold end	$T_{D}$	K	—	Conduction
heat current	$H$	W	—	Conduction
heat current of ground	$H_{g}$	W	—
thermal conductivity of ground	$K_{g}$	W m⁻¹ K⁻¹	—
cross section of ground	$A_{g}$	m²	—
hot end of ground	$T_{C,g}$	K	—
cold end of ground	$T_{D,g}$	K	—
length of the rod of ground	$L_{g}$	m	—
heat current of iron	$H_{i}$	W	—
thermal conductivity of iron	$K_{i}$	W m⁻¹ K⁻¹	—
cross section of iron	$A_{i}$	m²	—
hot end of iron	$T_{C,i}$	K	—
cold end of iron	$T_{D,i}$	K	—
length of the rod of iron	$L_{i}$	m	—
heat current of brass	$H_{b}$	W	—
thermal conductivity of brass	$K_{b}$	W m⁻¹ K⁻¹	—
cross section of brass	$A_{b}$	m²	—
hot end of brass	$T_{C,b}$	K	—
cold end of brass	$T_{D,b}$	K	—
length of the rod of brass	$L_{b}$	m	—

H_{r} = \frac{K_{r} \cdot A_{r} \cdot (T_{C,r} - T_{D,r})}{L_{r}}

heat flows down a rod

H_{g} = \frac{K_{g} \cdot A_{g} \cdot (T_{C,g} - T_{D,g})}{L_{g}}

heat flows down a rod(ground)

H_{i} = \frac{K_{i} \cdot A_{i} \cdot (T_{C,i} - T_{D,i})}{L_{i}}

heat flows down a rod(iron)

H_{b} = \frac{K_{b} \cdot A_{b} \cdot (T_{C,b} - T_{D,b})}{L_{b}}

heat flows down a rod(brass)

T_{C,b} = T_{D,i}

soldered end to end (hot end of brass)

H_{b} = H_{i}

soldered end to end (heat current of brass)