
The length of the two rods made up of the same metal and having the same area of the cross-section are 0.6m and 0.8m respectively. The temperature between the ends of the first rod is \[{90^0}C\] and \[{60^0}C\] and that for the other rod is \[{150^0}C\]and \[{110^0}C\]. For which rod the rate of conduction will be greater?
A. First
B. Second
C. Same for both
D. None of the above
Answer
164.4k+ 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.
Formula Used:
To find the rate of heat transfer the formula is,
\[\dfrac{Q}{t} = KA\dfrac{{\Delta T}}{L}\]
Where, A is a cross-sectional area, \[\Delta T\] is the temperature difference between two ends and L is the length of the cylinder.
Complete step by step solution:
Consider two rods that are made up of the same metal and have the same area of the cross-section \[{L_1} = 0.6m\] and \[{L_2} = 0.8m\] respectively. The temperature between the ends of the first rod is \[{T_1} = {90^0}C\] and \[{T_2} = {60^0}C\] and that for the second rod is \[{T_1} = {150^0}C\] and \[{T_2} = {110^0}C\]. We need to find for which rod the rate of conduction will be greater. To find the rate of conduction of heat the formula is,
\[\dfrac{Q}{t} = KA\dfrac{{\Delta T}}{L}\]
To find the rate of heat transfer for the first rod is,
\[\dfrac{{{Q_1}}}{t} = KA\dfrac{{\Delta T}}{{{L_1}}}\]………… (1)
Here,
\[\Delta T = {T_1} - {T_2}\]
\[ \Rightarrow \Delta T = 90 - 60\] and \[{L_1} = 0.6m\]
Substitute the value in equation (1) we get,
\[\dfrac{{{Q_1}}}{t} = KA\dfrac{{\left( {90 - 60} \right)}}{{0.6}}\]
\[\Rightarrow \dfrac{{{Q_1}}}{t} = 50KA\]
To find the rate of heat transfer for the first rod is,
\[\dfrac{{{Q_2}}}{t} = KA\dfrac{{\Delta T}}{{{L_2}}}\]………… (2)
Here, \[\Delta T = {T_1} - {T_2}\]
\[ \Rightarrow \Delta T = 150 - 110\] and \[{L_1} = 0.8m\]
Substitute the value in equation (2) we get,
\[\dfrac{{{Q_2}}}{t} = KA\dfrac{{\left( {150 - 110} \right)}}{{0.8}}\]
\[\therefore \dfrac{{{Q_1}}}{t} = 50KA\]
Since the rate of heat transfer for both rods is the same. Therefore, for both rods, the rate of conduction will be the same.
Hence, option D is the correct answer.
Note:In this given problem it is important to remember the formula to find the heat flow. Here we have considered 2 rods and need to find which rod conducts more. Heat transfer is the transport of heat (thermal energy) caused by temperature differences, as well as the resulting temperature distribution and variations. The study of transport phenomena is concerned with the transfer of momentum, energy, and mass via conduction, convection, and radiation.
Formula Used:
To find the rate of heat transfer the formula is,
\[\dfrac{Q}{t} = KA\dfrac{{\Delta T}}{L}\]
Where, A is a cross-sectional area, \[\Delta T\] is the temperature difference between two ends and L is the length of the cylinder.
Complete step by step solution:
Consider two rods that are made up of the same metal and have the same area of the cross-section \[{L_1} = 0.6m\] and \[{L_2} = 0.8m\] respectively. The temperature between the ends of the first rod is \[{T_1} = {90^0}C\] and \[{T_2} = {60^0}C\] and that for the second rod is \[{T_1} = {150^0}C\] and \[{T_2} = {110^0}C\]. We need to find for which rod the rate of conduction will be greater. To find the rate of conduction of heat the formula is,
\[\dfrac{Q}{t} = KA\dfrac{{\Delta T}}{L}\]
To find the rate of heat transfer for the first rod is,
\[\dfrac{{{Q_1}}}{t} = KA\dfrac{{\Delta T}}{{{L_1}}}\]………… (1)
Here,
\[\Delta T = {T_1} - {T_2}\]
\[ \Rightarrow \Delta T = 90 - 60\] and \[{L_1} = 0.6m\]
Substitute the value in equation (1) we get,
\[\dfrac{{{Q_1}}}{t} = KA\dfrac{{\left( {90 - 60} \right)}}{{0.6}}\]
\[\Rightarrow \dfrac{{{Q_1}}}{t} = 50KA\]
To find the rate of heat transfer for the first rod is,
\[\dfrac{{{Q_2}}}{t} = KA\dfrac{{\Delta T}}{{{L_2}}}\]………… (2)
Here, \[\Delta T = {T_1} - {T_2}\]
\[ \Rightarrow \Delta T = 150 - 110\] and \[{L_1} = 0.8m\]
Substitute the value in equation (2) we get,
\[\dfrac{{{Q_2}}}{t} = KA\dfrac{{\left( {150 - 110} \right)}}{{0.8}}\]
\[\therefore \dfrac{{{Q_1}}}{t} = 50KA\]
Since the rate of heat transfer for both rods is the same. Therefore, for both rods, the rate of conduction will be the same.
Hence, option D is the correct answer.
Note:In this given problem it is important to remember the formula to find the heat flow. Here we have considered 2 rods and need to find which rod conducts more. Heat transfer is the transport of heat (thermal energy) caused by temperature differences, as well as the resulting temperature distribution and variations. The study of transport phenomena is concerned with the transfer of momentum, energy, and mass via conduction, convection, and radiation.
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