Iron and Brass End to End

Q. An iron bar (length 0.1 m, cross-section 0.02 m², thermal conductivity 79 W m⁻¹ K⁻¹) and a brass bar (length 0.1 m, cross-section 0.02 m², thermal conductivity 109 W m⁻¹ K⁻¹) are soldered end to end. The free ends of the iron bar and brass bar are maintained at 373 K and 273 K respectively. What is the temperature of the junction of the two bars, in kelvin? expects 315 K

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. 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?

The playground

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}

The solution

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 Ruby

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

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

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

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)