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An iron bar of length 0.1m and area of cross section 0.02m*2 is joined length wise to a brass bar of the same length and diameter. Compute the rate of heat flows through the compound? Given k iron 79wm1K1 K brass= 109wm 1k1 the free ends of the iron bar and brass bar are maintained at 373k and 273k respectively.?
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An iron bar of length 0.1m and area of cross section 0.02m*2 is joined...
Introduction
In this problem, we are given an iron bar and a brass bar that are joined lengthwise. The iron bar has a length of 0.1m and an area of cross-section of 0.02m^2. We are given the thermal conductivities of iron and brass as 79 W/m·K and 109 W/m·K respectively. The free ends of the iron bar and brass bar are maintained at temperatures of 373K and 273K respectively. We need to calculate the rate of heat flow through the compound.
Formula
The rate of heat flow through a material is given by Fourier's Law:
Q = (k * A * ΔT) / L
Where:
Q - Rate of heat flow (in watts)
k - Thermal conductivity of the material (in W/m·K)
A - Area of cross-section (in m^2)
ΔT - Temperature difference (in K)
L - Length of the material (in m)
Calculations
We can calculate the rate of heat flow through the iron bar and the brass bar separately and then add them to get the total rate of heat flow through the compound.
For the iron bar:
k_iron = 79 W/m·K
A_iron = 0.02 m^2
ΔT_iron = 373K - 273K = 100K
L_iron = 0.1m
Using the formula, we can calculate the rate of heat flow through the iron bar:
Q_iron = (k_iron * A_iron * ΔT_iron) / L_iron
= (79 * 0.02 * 100) / 0.1
= 1580W
For the brass bar:
k_brass = 109 W/m·K
A_brass = 0.02 m^2
ΔT_brass = 273K - 373K = -100K (negative sign indicates temperature difference from hot to cold)
L_brass = 0.1m
Using the formula, we can calculate the rate of heat flow through the brass bar:
Q_brass = (k_brass * A_brass * ΔT_brass) / L_brass
= (109 * 0.02 * -100) / 0.1
= -2180W
Total rate of heat flow through the compound:
Q_total = Q_iron + Q_brass
= 1580W + (-2180W)
= -600W
Explanation
The negative sign in the rate of heat flow through the brass bar indicates that heat is flowing from the brass bar to the iron bar. This is because the brass bar is at a lower temperature than the iron bar. The rate of heat flow through the compound is -600W, which means that heat is flowing from the iron bar to the brass bar at a rate of 600W.
In this problem, we are given an iron bar and a brass bar that are joined lengthwise. The iron bar has a length of 0.1m and an area of cross-section of 0.02m^2. We are given the thermal conductivities of iron and brass as 79 W/m·K and 109 W/m·K respectively. The free ends of the iron bar and brass bar are maintained at temperatures of 373K and 273K respectively. We need to calculate the rate of heat flow through the compound.
Formula
The rate of heat flow through a material is given by Fourier's Law:
Q = (k * A * ΔT) / L
Where:
Q - Rate of heat flow (in watts)
k - Thermal conductivity of the material (in W/m·K)
A - Area of cross-section (in m^2)
ΔT - Temperature difference (in K)
L - Length of the material (in m)
Calculations
We can calculate the rate of heat flow through the iron bar and the brass bar separately and then add them to get the total rate of heat flow through the compound.
For the iron bar:
k_iron = 79 W/m·K
A_iron = 0.02 m^2
ΔT_iron = 373K - 273K = 100K
L_iron = 0.1m
Using the formula, we can calculate the rate of heat flow through the iron bar:
Q_iron = (k_iron * A_iron * ΔT_iron) / L_iron
= (79 * 0.02 * 100) / 0.1
= 1580W
For the brass bar:
k_brass = 109 W/m·K
A_brass = 0.02 m^2
ΔT_brass = 273K - 373K = -100K (negative sign indicates temperature difference from hot to cold)
L_brass = 0.1m
Using the formula, we can calculate the rate of heat flow through the brass bar:
Q_brass = (k_brass * A_brass * ΔT_brass) / L_brass
= (109 * 0.02 * -100) / 0.1
= -2180W
Total rate of heat flow through the compound:
Q_total = Q_iron + Q_brass
= 1580W + (-2180W)
= -600W
Explanation
The negative sign in the rate of heat flow through the brass bar indicates that heat is flowing from the brass bar to the iron bar. This is because the brass bar is at a lower temperature than the iron bar. The rate of heat flow through the compound is -600W, which means that heat is flowing from the iron bar to the brass bar at a rate of 600W.
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An iron bar of length 0.1m and area of cross section 0.02m*2 is joined length wise to a brass bar of the same length and diameter. Compute the rate of heat flows through the compound? Given k iron 79wm1K1 K brass= 109wm 1k1 the free ends of the iron bar and brass bar are maintained at 373k and 273k respectively.?
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An iron bar of length 0.1m and area of cross section 0.02m*2 is joined length wise to a brass bar of the same length and diameter. Compute the rate of heat flows through the compound? Given k iron 79wm1K1 K brass= 109wm 1k1 the free ends of the iron bar and brass bar are maintained at 373k and 273k respectively.? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about An iron bar of length 0.1m and area of cross section 0.02m*2 is joined length wise to a brass bar of the same length and diameter. Compute the rate of heat flows through the compound? Given k iron 79wm1K1 K brass= 109wm 1k1 the free ends of the iron bar and brass bar are maintained at 373k and 273k respectively.? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for An iron bar of length 0.1m and area of cross section 0.02m*2 is joined length wise to a brass bar of the same length and diameter. Compute the rate of heat flows through the compound? Given k iron 79wm1K1 K brass= 109wm 1k1 the free ends of the iron bar and brass bar are maintained at 373k and 273k respectively.?.
An iron bar of length 0.1m and area of cross section 0.02m*2 is joined length wise to a brass bar of the same length and diameter. Compute the rate of heat flows through the compound? Given k iron 79wm1K1 K brass= 109wm 1k1 the free ends of the iron bar and brass bar are maintained at 373k and 273k respectively.? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about An iron bar of length 0.1m and area of cross section 0.02m*2 is joined length wise to a brass bar of the same length and diameter. Compute the rate of heat flows through the compound? Given k iron 79wm1K1 K brass= 109wm 1k1 the free ends of the iron bar and brass bar are maintained at 373k and 273k respectively.? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for An iron bar of length 0.1m and area of cross section 0.02m*2 is joined length wise to a brass bar of the same length and diameter. Compute the rate of heat flows through the compound? Given k iron 79wm1K1 K brass= 109wm 1k1 the free ends of the iron bar and brass bar are maintained at 373k and 273k respectively.?.
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An iron bar of length 0.1m and area of cross section 0.02m*2 is joined length wise to a brass bar of the same length and diameter. Compute the rate of heat flows through the compound? Given k iron 79wm1K1 K brass= 109wm 1k1 the free ends of the iron bar and brass bar are maintained at 373k and 273k respectively.?, a detailed solution for An iron bar of length 0.1m and area of cross section 0.02m*2 is joined length wise to a brass bar of the same length and diameter. Compute the rate of heat flows through the compound? Given k iron 79wm1K1 K brass= 109wm 1k1 the free ends of the iron bar and brass bar are maintained at 373k and 273k respectively.? has been provided alongside types of An iron bar of length 0.1m and area of cross section 0.02m*2 is joined length wise to a brass bar of the same length and diameter. Compute the rate of heat flows through the compound? Given k iron 79wm1K1 K brass= 109wm 1k1 the free ends of the iron bar and brass bar are maintained at 373k and 273k respectively.? theory, EduRev gives you an
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