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Two bars of same length and same cross -sectional area but of

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Two bars of same length and same cross -sectional area but of different thermal conductivities $K_{1}$ and $K_{2}$ are joined end to end as shown in the figure .One end of the compound bar is at temperature $T_{1}$ and the opposite end at temperature $T_{2}$ $(w h ere T_{1} > T_{2})$ .
The temperature of the junction is

\frac{K _{1} T _{1} + K _{2} T _{2}}{K _{1} + K _{2}}

\frac{K _{1} T _{2} + K _{2} T _{1}}{K _{1} + K _{2}}

\frac{K _{1} ( T _{1} + T _{2} )}{K _{2}}

\frac{K _{2} ( T _{1} + T _{2} )}{K _{1}}

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Two bars of same length and same cross -sectional area but of different thermal conductivities

K_{1}

and

K_{2}

are joined end to end as shown in the figure .One end of the compound bar is at temperature

T_{1}

and the opposite end at temperature

T_{2}

(w h ere T_{1} > T_{2})

.
The temperature of the junction is

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Text solutionVerified

Let

L

and

A

be length and area of cross-section of each bar respectively.

∴

Heat current through the bar 1 is

H_{1} = \frac{K _{1} A ( T _{1} - T _{0} )}{L}

At steady state,

H_{1} = H_{2}

∴ \frac{K _{1} A ( T _{1} - T _{0} )}{L} = \frac{K _{2} A ( T _{0} - T _{2} )}{L}

K_{1} (T_{1} - T_{0}) = K_{2} (T_{0} - T_{2})

K_{1} T_{1} - K_{1} T_{0} = K_{2} T_{0} - K_{2} T_{2}

K_{1} T_{0} + K_{2} T_{0} = K_{1} T_{1} + K_{2} T_{2}

T_{0} (K_{1} + K_{2}) = K_{1} T_{1} + K_{2} T_{2}

T_{0} = \frac{K _{1} T _{1} + K _{2} T _{2}}{( K _{1} + K _{2} )}

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Question Text	Two bars of same length and same cross -sectional area but of different thermal conductivities $K_{1}$ and $K_{2}$ are joined end to end as shown in the figure .One end of the compound bar is at temperature $T_{1}$ and the opposite end at temperature $T_{2}$ $(w h ere T_{1} > T_{2})$ . The temperature of the junction is
Answer Type	Text solution:1
Upvotes	150