Question

A long rod, insulated to prevent heat loss along its sides, is in perfect thermal contact with boiling water (at atmospheric pressure) at one end and with an ice-water mixture at the other. The rod consists of a 1.00 m section of copper (with one end in the boiling water) joined end-to-end to a length  L₂ of steel (with one end in the ice water). Both sections of the rod have cross-sectional areas of 4.00 cm². The temperature of the copper-steel junction is 65.0 ℃ after a steady-state has been reached. Assume that the thermal conductivities of copper and steel are given by $$ k_{\text {copper }}=385 \frac{\mathrm{W}}{\mathrm{m} \cdot \mathrm{K}} \text { and } k_{\text {steel }}=50.2 \frac{\mathrm{W}}{\mathrm{m} \cdot \mathrm{K}} $$.

How much heat per second H flows from the boiling water to the ice-water mixture?

Answer 1

Concepts and reason

The concept required to solve the given problem is thermodynamics.

Initially, calculate the amount of heat per second flows from the steam bath to the ice water mixture using the relation between heat and temperature. Next, write the relation between length and temperature. Finally, calculate the length of the steel section by using the relation between length and temperature.

Fundamentals

The expression for the amount of heat is,

$Q = \frac{{kA\Delta Tt}}{L}$

Here, k is the thermal conductivity of the material, A is the cross sectional area, $\Delta T$ is the change in temperature, L is the thickness of the material, and t is the time taken.

The amount of heat transferred per second is,

$H = \frac{{kA\Delta T}}{L}$

Substitute ${T_{\rm{H}}} - {T_{\rm{C}}}$ for $\Delta T$ in the above equation.

$H = \frac{{kA\left( {{T_{\rm{H}}} - {T_{\rm{C}}}} \right)}}{L}$

Substitute $385\;{\rm{W/m}}{\rm{.K}}$ for k, $4\;{\rm{c}}{{\rm{m}}^2}$ for A, $100^\circ {\rm{C}}$ for ${T_{\rm{H}}}$, $65^\circ {\rm{C}}$ for ${T_{\rm{C}}}$, and 1 m for L in the above equation.

$\begin{array}{c}\\H = \frac{{\left( {385\;{\rm{W/m}}{\rm{.K}}} \right)\left( {4\;{\rm{c}}{{\rm{m}}^2}} \right)\left( {\frac{{{{10}^{ - 4}}\;{{\rm{m}}^2}}}{{{\rm{c}}{{\rm{m}}^{\rm{2}}}}}} \right)\left( {\left( {100 + 273} \right)\;{\rm{K}} - \left( {65 + 273} \right)\;{\rm{K}}} \right)}}{{1\;{\rm{m}}}}\\\\ = 5.39\;{\rm{W}}\\\end{array}$

The length of the rod is,

$\begin{array}{c}\\{L_2} = {L_1}\left( {\frac{{{k_2}}}{{{k_1}}}} \right)\left( {\frac{{\Delta {T_2}}}{{\Delta {T_1}}}} \right)\\\\ = {L_1}\left( {\frac{{{k_2}}}{{{k_1}}}} \right)\left( {\frac{{{T_{\rm{H}}} - {T_{\rm{C}}}}}{{{T_{\rm{C}}}}}} \right)\\\end{array}$

Substitute $50.2\;{\rm{W/m}}{\rm{.K}}$ for${k_2}$, $385\;{\rm{W/m}}{\rm{.K}}$for${k_1}$, $100^\circ {\rm{C}}$ for${T_{\rm{H}}}$, $35^\circ {\rm{C}}$ for${T_{\rm{C}}}$, and 1 m for L in the above equation.

$\begin{array}{c}\\{L_2} = \left( {1\;{\rm{m}}} \right)\left( {\frac{{50.2\;{\rm{W/m}}{\rm{.K}}}}{{385\;{\rm{W/m}}{\rm{.K}}}}} \right)\left( {\frac{{\left( {\left( {100 + 273} \right)\;{\rm{K}} - \left( {35 + 273} \right)\;{\rm{K}}} \right)}}{{\left( {35 + 273} \right)\;{\rm{K}}}}} \right)\\\\ = \left( {1\;{\rm{m}}} \right)\left( {\frac{{50.2\;{\rm{W/m}}{\rm{.K}}}}{{385\;{\rm{W/m}}{\rm{.K}}}}} \right)\left( {\frac{{373\;{\rm{K}} - 308\;{\rm{K}}}}{{308\;{\rm{K}}}}} \right)\\\\ = 0.242\;{\rm{m}}\\\end{array}$

Ans:

The heat per second flows from the steam bath to the ice water mixture is 5.39 W.

The length of the steel section is 0.242 m.

Answer 2

Heat flow rate through Cu = K_1 A delta T/l_1 = 385 times 4 times 10^-4 times (100 - 65)/1 = 5.39 wolt = 5.39 joule per second (b) At steady state K_2 A (65 - 0)/L_2 = K_1 A(100 - 65)/1 50.2 times 65/L_2 = 385 times 35 L_2 = 0.242m