
Two rectangular blocks A and B of different metals have the same length and same area of cross-section. They have kept in such a way that their cross-sectional area touches each other. The temperature at one end of A is \[{100^0}C\] and that of B at the other end is \[{0^0}C\]. If the ratio of their thermal conductivity is 1:3, then under a steady state, find the temperature of the junction in contact.
A. \[{25^0}C\]
B. \[{50^0}C\]
C. \[{75^0}C\]
D. \[{100^0}C\]
Answer
161.1k+ views
Hint:In order to solve this problem we need to understand the rate of heat transfer. The rate of flow of heat is the amount of heat that is transferred per unit of time. Here, using the formula for heat flow we are going to find the solution. When the two rods are kept in contact with each other then, the temperature at the junction of 2 rods is known as junction temperature.
Formula Used:
To find the temperature of the junction in contact the formula is,
\[\theta = \dfrac{{{K_1}{\theta _1} + {K_2}{\theta _2}}}{{{K_1} + {K_2}}}\]
Where, K is the thermal conductivity and \[{\theta _1},{\theta _2}\] is the temperature of a rectangular bar.
Complete step by step solution:
If two rectangular blocks A and B of different metals have the same length and same area of cross-section then, the ratio of their thermal conductivity is \[\dfrac{{{K_1}}}{{{K_2}}} = \dfrac{1}{3}\].
We need to find the temperature of the junction in contact.
If \[\dfrac{{{K_1}}}{{{K_2}}} = \dfrac{1}{3}\] then, \[{K_2} = 3{K_1}\]
The temperature of the junction in contact with the formula is,
\[\theta = \dfrac{{{K_1}{\theta _1} + {K_2}{\theta _2}}}{{{K_1} + {K_2}}}\]
Here, \[{K_2} = 3{K_1}\], \[{\theta _1} = {100^0}C\] and \[{\theta _1} = {0^0}C\]
Substitute the value in above equation we obtain,
\[\theta = \dfrac{{1 \times 100{K_1} + 3{K_1} \times 0}}{{{K_1} + 3{K_1}}} \\ \]
\[\Rightarrow \theta = \dfrac{{{K_1}\left( {1 \times 100 + 3 \times 0} \right)}}{{{K_1}\left( {1 + 3} \right)}} \\ \]
\[\Rightarrow \theta = \dfrac{{\left( {1 \times 100 + 3 \times 0} \right)}}{4} \\ \]
\[\Rightarrow \theta = \dfrac{{100}}{4} \\ \]
\[\therefore \theta = {25^0}C\]
Therefore, the temperature of the junction in contact is \[{25^0}C\].
Hence, option A is the correct answer.
Note: Here in the given problem it is important to remember the equation for the temperature of the junction. Using the formula, we can easily find the solution.
Formula Used:
To find the temperature of the junction in contact the formula is,
\[\theta = \dfrac{{{K_1}{\theta _1} + {K_2}{\theta _2}}}{{{K_1} + {K_2}}}\]
Where, K is the thermal conductivity and \[{\theta _1},{\theta _2}\] is the temperature of a rectangular bar.
Complete step by step solution:
If two rectangular blocks A and B of different metals have the same length and same area of cross-section then, the ratio of their thermal conductivity is \[\dfrac{{{K_1}}}{{{K_2}}} = \dfrac{1}{3}\].
We need to find the temperature of the junction in contact.
If \[\dfrac{{{K_1}}}{{{K_2}}} = \dfrac{1}{3}\] then, \[{K_2} = 3{K_1}\]
The temperature of the junction in contact with the formula is,
\[\theta = \dfrac{{{K_1}{\theta _1} + {K_2}{\theta _2}}}{{{K_1} + {K_2}}}\]
Here, \[{K_2} = 3{K_1}\], \[{\theta _1} = {100^0}C\] and \[{\theta _1} = {0^0}C\]
Substitute the value in above equation we obtain,
\[\theta = \dfrac{{1 \times 100{K_1} + 3{K_1} \times 0}}{{{K_1} + 3{K_1}}} \\ \]
\[\Rightarrow \theta = \dfrac{{{K_1}\left( {1 \times 100 + 3 \times 0} \right)}}{{{K_1}\left( {1 + 3} \right)}} \\ \]
\[\Rightarrow \theta = \dfrac{{\left( {1 \times 100 + 3 \times 0} \right)}}{4} \\ \]
\[\Rightarrow \theta = \dfrac{{100}}{4} \\ \]
\[\therefore \theta = {25^0}C\]
Therefore, the temperature of the junction in contact is \[{25^0}C\].
Hence, option A is the correct answer.
Note: Here in the given problem it is important to remember the equation for the temperature of the junction. Using the formula, we can easily find the solution.
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