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Mud houses are cooler in summer and warmer in winter because
Mud houses do not get warmer in summer because mud is a bad conductor of heat. They resist the movement of heat that would make its interior warm with hot weather and cold in winter by either preventing the movement of heat within or outside in. Click here for more: https://www.tiwariacademy.com/nceRead more
Mud houses do not get warmer in summer because mud is a bad conductor of heat. They resist the movement of heat that would make its interior warm with hot weather and cold in winter by either preventing the movement of heat within or outside in.
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A body cools from 80 °C to 64 °C in 5 minutes and same body cools from 80 °C to 52 °C in 10 minutes. What is the temperature of surrounding?
We can solve this using Newton's Law of Cooling, which states, (dT/dt) = -k(T - Tₛ) (dT/dt) is the rate of change of temperature k is a constant T is the temperature of the body Tₛ is the temperature of the surroundings. We are given that the body cools from 80°C to 64°C in 5 minutes and again fromRead more
We can solve this using Newton’s Law of Cooling, which states,
(dT/dt) = -k(T – Tₛ)
(dT/dt) is the rate of change of temperature
k is a constant
T is the temperature of the body
Tₛ is the temperature of the surroundings.
We are given that the body cools from 80°C to 64°C in 5 minutes and again from 80°C to 52°C in 10 minutes. Applying Newton’s Law, we can derive the temperature of the surroundings.
We can use this formula to find the temperatures:
ln[(T₁ – Tₛ) / (T₂ – Tₛ)] = k(t₂ – t₁)
For the first cooling process, where it cools from 80 °C to 64 °C in 5 minutes:
ln[(80 – Tₛ) / (64 – Tₛ)] = k × 5
For the second cooling (from 80 °C to 52 °C in 10 minutes):
ln[(80 – Tₛ) / (52 – Tₛ)] = k × 10
This system of equations solves for the value of Tₛ.
After solving, we find that the temperature of the surroundings is 25 °C.
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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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