We can solve this problem using the formula for heat transfer through a rod, which is as follows: Q = (kA(T₁ - T₂)) / L where: - Q is the heat transferred, - k is the thermal conductivity, - A is the cross-sectional area, - (T₁ - T₂) is the temperature difference, - L is the length of the rod. SinceRead more
We can solve this problem using the formula for heat transfer through a rod, which is as follows:
Q = (kA(T₁ – T₂)) / L
where:
– Q is the heat transferred,
– k is the thermal conductivity,
– A is the cross-sectional area,
– (T₁ – T₂) is the temperature difference,
– L is the length of the rod.
Since the rods are connected end to end, heat transfer by both the rods is identical. The heat transfer by both the rods is the same:
(k₁A(T₁ – T)) / L = (k₂A(T – T₂)) / L
where:
– k₁ and k₂ are the thermal conductivities of the first and second rods,
– T₁ and T₂ are the temperatures at the free ends of the first and second rods
– T is the temperature at the junction.
Given:
– The ratio of thermal conductivity is k₁ : k₂ = 5 : 3,
– T₁ = 100°C,
– T₂ = 20°C.
The thermal resistance (R) of a rod is given as: R = L / (kA) where: L is the length of the rod, k is the thermal conductivity, A is the cross-sectional area. Given two rods made of different materials, having equal thermal resistance: R₁ = R₂, and the area of cross-section: A₁ = A₂, we obtain: L₁ /Read more
The thermal resistance (R) of a rod is given as:
R = L / (kA)
where:
L is the length of the rod,
k is the thermal conductivity,
A is the cross-sectional area.
Given two rods made of different materials, having equal thermal resistance: R₁ = R₂, and the area of cross-section: A₁ = A₂, we obtain:
L₁ / (k₁A) = L₂ / (k₂A)
Area of cross-section cancels out:
L₁ / k₁ = L₂ / k₂
Rearranging to get the ratio of lengths:
L₁ / L₂ = k₁ / k₂
k₁ : k₂ = 5 : 4
L₁ : L₂ = 4 : 5
Answer: 4 : 5
We have to calculate the heat capacity of both bodies and compare the initial temperatures to determine which way the heat will flow. The formula for heat capacity (C) is: C = m × s where: m = mass, s = specific heat. Body A: Cₐ = 0.5 × 0.85 = 0.425 Body B: C_b = 0.3 × 0.9 = 0.27 The body having a hRead more
We have to calculate the heat capacity of both bodies and compare the initial temperatures to determine which way the heat will flow.
The formula for heat capacity (C) is:
C = m × s
where:
m = mass,
s = specific heat.
Body A:
Cₐ = 0.5 × 0.85 = 0.425
Body B:
C_b = 0.3 × 0.9 = 0.27
The body having a higher product of mass and specific heat will have more thermal energy at the same temperature. However, in this case, the temperature plays a major role in deciding where the direction of heat will be.
Initial temperatures:
– A = 60°C
– B = 90°C
As body B has a higher temperature than body A, the heat will flow from B to A.
To cause heat to move from one end of a solid to another, there must exist a temperature gradient. This would mean that two parts of the solid must differ in temperature as heat flows from the region with higher temperature towards the region of lower temperature. Click here for more: https://www.tiRead more
To cause heat to move from one end of a solid to another, there must exist a temperature gradient. This would mean that two parts of the solid must differ in temperature as heat flows from the region with higher temperature towards the region of lower temperature.
The energy radiated by a body is given by the Stefan-Boltzmann law: E ∝ T⁴ where E is the energy emitted and T is the temperature in Kelvin. To find the ratio of the energies emitted at two temperatures, we use the formula: (E₂ / E₁) = (T₂ / T₁)⁴ First, convert the temperatures from Celsius to KelviRead more
The energy radiated by a body is given by the Stefan-Boltzmann law:
E ∝ T⁴
where E is the energy emitted and T is the temperature in Kelvin.
To find the ratio of the energies emitted at two temperatures, we use the formula:
(E₂ / E₁) = (T₂ / T₁)⁴
First, convert the temperatures from Celsius to Kelvin:
T₁ = 27 + 273 = 300 K T₂ = 92.7 + 273 = 365.7 K
Two rods having thermal conductivity in the ratio of 5 : 3 having equal lengths and equal cross-sectional area are joined end to end. If the temperature of the free end of the first rod is 100°C and free end of the second rod is 20°C, then the temperature of the jucntion is
We can solve this problem using the formula for heat transfer through a rod, which is as follows: Q = (kA(T₁ - T₂)) / L where: - Q is the heat transferred, - k is the thermal conductivity, - A is the cross-sectional area, - (T₁ - T₂) is the temperature difference, - L is the length of the rod. SinceRead more
We can solve this problem using the formula for heat transfer through a rod, which is as follows:
Q = (kA(T₁ – T₂)) / L
where:
– Q is the heat transferred,
– k is the thermal conductivity,
– A is the cross-sectional area,
– (T₁ – T₂) is the temperature difference,
– L is the length of the rod.
Since the rods are connected end to end, heat transfer by both the rods is identical. The heat transfer by both the rods is the same:
(k₁A(T₁ – T)) / L = (k₂A(T – T₂)) / L
where:
– k₁ and k₂ are the thermal conductivities of the first and second rods,
– T₁ and T₂ are the temperatures at the free ends of the first and second rods
– T is the temperature at the junction.
Given:
– The ratio of thermal conductivity is k₁ : k₂ = 5 : 3,
– T₁ = 100°C,
– T₂ = 20°C.
Now, equating the heat transfer in both rods:
(5(100 – T)) / 3 = (T – 20)
Solving for T:
5(100 – T) = 3(T – 20)
500 – 5T = 3T – 60
500 + 60 = 8T
560 = 8T
T = 70°C
Hence, the temperature at the junction is 70°C.
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The ratio of thermal conductivity of two rods of different material is 5 : 4. The two rods of same area of cross-section and same thermal resistance will have the lengths in the ratio
The thermal resistance (R) of a rod is given as: R = L / (kA) where: L is the length of the rod, k is the thermal conductivity, A is the cross-sectional area. Given two rods made of different materials, having equal thermal resistance: R₁ = R₂, and the area of cross-section: A₁ = A₂, we obtain: L₁ /Read more
The thermal resistance (R) of a rod is given as:
R = L / (kA)
where:
L is the length of the rod,
k is the thermal conductivity,
A is the cross-sectional area.
Given two rods made of different materials, having equal thermal resistance: R₁ = R₂, and the area of cross-section: A₁ = A₂, we obtain:
L₁ / (k₁A) = L₂ / (k₂A)
Area of cross-section cancels out:
L₁ / k₁ = L₂ / k₂
Rearranging to get the ratio of lengths:
L₁ / L₂ = k₁ / k₂
k₁ : k₂ = 5 : 4
L₁ : L₂ = 4 : 5
Answer: 4 : 5
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A body A of mass 0.5 kg and specific heat 0.85 is at a temperature of 60°C. Another body B of mass 0.3 kg and specific heat 0.9 is at a temperature of 90°C. When they are connected to a conducting rod, heat will flow from
We have to calculate the heat capacity of both bodies and compare the initial temperatures to determine which way the heat will flow. The formula for heat capacity (C) is: C = m × s where: m = mass, s = specific heat. Body A: Cₐ = 0.5 × 0.85 = 0.425 Body B: C_b = 0.3 × 0.9 = 0.27 The body having a hRead more
We have to calculate the heat capacity of both bodies and compare the initial temperatures to determine which way the heat will flow.
The formula for heat capacity (C) is:
C = m × s
where:
m = mass,
s = specific heat.
Body A:
Cₐ = 0.5 × 0.85 = 0.425
Body B:
C_b = 0.3 × 0.9 = 0.27
The body having a higher product of mass and specific heat will have more thermal energy at the same temperature. However, in this case, the temperature plays a major role in deciding where the direction of heat will be.
Initial temperatures:
– A = 60°C
– B = 90°C
As body B has a higher temperature than body A, the heat will flow from B to A.
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In order that the heat flows from one part of a solid to another part, what is required?
To cause heat to move from one end of a solid to another, there must exist a temperature gradient. This would mean that two parts of the solid must differ in temperature as heat flows from the region with higher temperature towards the region of lower temperature. Click here for more: https://www.tiRead more
To cause heat to move from one end of a solid to another, there must exist a temperature gradient. This would mean that two parts of the solid must differ in temperature as heat flows from the region with higher temperature towards the region of lower temperature.
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See lesshttps://www.tiwariacademy.com/ncert-solutions/class-11/physics/chapter-10/
If a body is heated from 27° C to 92.7 °C, then the ratio of their energies of radiations emitted will be
The energy radiated by a body is given by the Stefan-Boltzmann law: E ∝ T⁴ where E is the energy emitted and T is the temperature in Kelvin. To find the ratio of the energies emitted at two temperatures, we use the formula: (E₂ / E₁) = (T₂ / T₁)⁴ First, convert the temperatures from Celsius to KelviRead more
The energy radiated by a body is given by the Stefan-Boltzmann law:
E ∝ T⁴
where E is the energy emitted and T is the temperature in Kelvin.
To find the ratio of the energies emitted at two temperatures, we use the formula:
(E₂ / E₁) = (T₂ / T₁)⁴
First, convert the temperatures from Celsius to Kelvin:
T₁ = 27 + 273 = 300 K T₂ = 92.7 + 273 = 365.7 K
Now, find the ratio:
(E₂ / E₁) = (365.7 / 300)⁴ ≈ (1.219)⁴ ≈ 2.1⁴ ≈ 16
So, the ratio of the energies emitted is 1 : 16.
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