
One end of a metal rod of length 1.0m and area of cross-section \[100\,c{m^2}\]is maintained \[{100^0}C\]. If the other end of the rod is maintained at \[{0^0}C\] then find the quantity of heat transmitted through the rod per minute. (Coefficient of thermal conductivity of metal rod \[ = 100\,W/kg/k\] )
A. \[3 \times {10^3}J\]
B. \[6 \times {10^3}J\]
C. \[9 \times {10^3}J\]
D. \[12 \times {10^3}J\]
Answer
164.1k+ 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 temperature. A temperature gradient is defined as the change in temperature over a specified distance between two isothermal surfaces or two points in a material.
Formula Used:
To find the heat flow the formula is,
\[\dfrac{Q}{t} = KA\dfrac{{\Delta T}}{L}\]
Where, A is a cross-sectional area of metal rod, \[\Delta T\] is temperature difference between two ends of the metal rod, L is length of the metal rod and K is thermal conductivity.
Complete step by step solution:
Here, we have a metal rod of length 1.0m and area of cross-section \[100c{m^2}\]that is maintained \[{100^0}C\]and the other end of the rod at \[{0^0}C\]then we need to find the quantity of heat that is transmitted through the rod per minute. We have the formula to find the heat flow in a metal rod that is,
\[\dfrac{Q}{t} = KA\dfrac{{\Delta T}}{L}\]
Given, \[K = 100W/kg/k\], \[A = 100c{m^{^2}}\], \[L = 1.0m\] and \[\Delta T = \left( {{{100}^0}C - {0^0}C} \right)\]
Substitute the value in the above equation we obtain,
\[\dfrac{Q}{t} = \dfrac{{100 \times 100 \times {{10}^{ - 4}} \times \left( {{{100}^0}C - {0^0}C} \right)}}{1}\]
\[\Rightarrow \dfrac{Q}{t} = 1000000 \times {10^{ - 4}}\]
\[\Rightarrow \dfrac{Q}{t} = 100\]
The quantity of heat that is transmitted through the rod per minute is,
\[\dfrac{Q}{t} = 100 \times 60\]
\[\Rightarrow \dfrac{Q}{t} = 6000\]
\[\therefore \dfrac{Q}{t} = 6 \times {10^3}J\]
Therefore, the quantity of heat transmitted through the rod per minute is \[6 \times {10^3}J\].
Hence, option B is the correct answer.
Note: Here in the given problem it is important to remember that the equation for the flow of heat in the rod and when we want to find the quantity of heat transmitted through the rod per minute, we have to multiply the obtained answer by 60sec (per minute).
Formula Used:
To find the heat flow the formula is,
\[\dfrac{Q}{t} = KA\dfrac{{\Delta T}}{L}\]
Where, A is a cross-sectional area of metal rod, \[\Delta T\] is temperature difference between two ends of the metal rod, L is length of the metal rod and K is thermal conductivity.
Complete step by step solution:
Here, we have a metal rod of length 1.0m and area of cross-section \[100c{m^2}\]that is maintained \[{100^0}C\]and the other end of the rod at \[{0^0}C\]then we need to find the quantity of heat that is transmitted through the rod per minute. We have the formula to find the heat flow in a metal rod that is,
\[\dfrac{Q}{t} = KA\dfrac{{\Delta T}}{L}\]
Given, \[K = 100W/kg/k\], \[A = 100c{m^{^2}}\], \[L = 1.0m\] and \[\Delta T = \left( {{{100}^0}C - {0^0}C} \right)\]
Substitute the value in the above equation we obtain,
\[\dfrac{Q}{t} = \dfrac{{100 \times 100 \times {{10}^{ - 4}} \times \left( {{{100}^0}C - {0^0}C} \right)}}{1}\]
\[\Rightarrow \dfrac{Q}{t} = 1000000 \times {10^{ - 4}}\]
\[\Rightarrow \dfrac{Q}{t} = 100\]
The quantity of heat that is transmitted through the rod per minute is,
\[\dfrac{Q}{t} = 100 \times 60\]
\[\Rightarrow \dfrac{Q}{t} = 6000\]
\[\therefore \dfrac{Q}{t} = 6 \times {10^3}J\]
Therefore, the quantity of heat transmitted through the rod per minute is \[6 \times {10^3}J\].
Hence, option B is the correct answer.
Note: Here in the given problem it is important to remember that the equation for the flow of heat in the rod and when we want to find the quantity of heat transmitted through the rod per minute, we have to multiply the obtained answer by 60sec (per minute).
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