According to the Stefan-Boltzmann law, the energy radiated per second by a body will be determined by how much power an object is radiating. And this shall be proportional to its fourth power of its temperature and its surface area. Therefore the formula describing power radiated is P = σ A T⁴ WhereRead more
According to the Stefan-Boltzmann law, the energy radiated per second by a body will be determined by how much power an object is radiating. And this shall be proportional to its fourth power of its temperature and its surface area. Therefore the formula describing power radiated is
P = σ A T⁴
Where,
– P is the power radiated,
– σ is the Stefan-Boltzmann constant,
– A is the surface area of the sphere
– T is the temperature of the sphere.
For a sphere, the surface area A is given by:
A = 4 π r²
Now, let’s calculate the energy radiated by both spheres:
For the first sphere:
– Radius r₁ = 1 m,
– Temperature T₁ = 4000 K.
Surface area:
A₁ = 4 π (1)² = 4 π m²
The power radiated:
P₁ = σ A₁ T₁⁴ = σ · 4 π · (4000)⁴
For the second sphere:
– Radius r₂ = 4 m,
– Temperature T₂ = 2000 K.
Surface area:
A₂ = 4 π (4)² = 64 π m²
The power radiated:
P₂ = σ A₂ T₂⁴ = σ · 64 π · (2000)⁴
Now, comparing P₁ and P₂:
– P₁ has a small surface area but a much greater temperature.
– P₂ has a larger surface area but a lower temperature.
However, the temperature factor dominates as the power varies as the fourth power of the temperature. So, 4000⁴ will be much larger than 2000⁴.
Hence, even though the surface area of the first sphere is smaller, the energy radiated per second by the first sphere will be greater than that by the second.
Conclusion:
The energy radiated per second by the first sphere is greater than that by the second.
We can use Wien's Law to solve this problem. Wien's Law relates the temperature of a black body to the wavelength at which it emits maximum radiation. The formula is: λₘₐₓ T = b Where: - λₘₐₓ is the wavelength at which the maximum intensity occurs, - T is the temperature of the black body, - b is WiRead more
We can use Wien’s Law to solve this problem. Wien’s Law relates the temperature of a black body to the wavelength at which it emits maximum radiation. The formula is:
λₘₐₓ T = b
Where:
– λₘₐₓ is the wavelength at which the maximum intensity occurs,
– T is the temperature of the black body,
– b is Wien’s constant, which is approximately 2.898 × 10⁶ nm·K.
Let’s denote the temperatures of the Sun and the North Star as Tₛᵤₙ and Tₙₒᵣₜₕₛₜₐᵣ, and their corresponding maximum wavelengths as λₛᵤₙ and λₙₒᵣₜₕₛₜₐᵣ.
For the Sun:
λₛᵤₙ = 510 nm
For the North Star:
λₙₒᵣₜₕₛₜₐᵣ = 350 nm
Applying Wien’s Law for both stars, we can write:
λₛᵤₙ Tₛᵤₙ = λₙₒᵣₜₕₛₜₐᵣ Tₙₒᵣₜₕₛₜₐᵣ
Now, solving for the ratio of their temperatures:
Tₛᵤₙ / Tₙₒᵣₜₕₛₜₐᵣ = λₙₒᵣₜₕₛₜₐᵣ / λₛᵤₙ
Substituting the values:
Tₛᵤₙ / Tₙₒᵣₜₕₛₜₐᵣ = 350 / 510 ≈ 0.686
Therefore, the ratio of surface temperatures of Sun and North Star is approximately 0.69.
We can solve for this using Stefan-Boltzmann law, which relates the power radiated by a black body with the following relationship: P = σ A T⁴ where, P = power radiated σ = Stefan-Boltzmann constant A = surface area of the sphere T = temperature of the sphere Now, the surface area of the sphere is gRead more
We can solve for this using Stefan-Boltzmann law, which relates the power radiated by a black body with the following relationship:
P = σ A T⁴
where,
P = power radiated
σ = Stefan-Boltzmann constant
A = surface area of the sphere
T = temperature of the sphere
Now, the surface area of the sphere is given as
A = 4 π r²
Initially,
– The radius of the sphere is r = 12 cm = 0.12 m,
– The temperature is T = 500 K,
– The power radiated is P = 450 W.
Let us first calculate how much power is given out initially using the Stefan-Boltzmann law:
P₁ = σ A₁ T₁⁴
Now, when the radius is halved or become r₂ = 0.06 m and the temperature is doubled or become T₂ = 1000 K, then the new power will be as follows:
P₂ = σ A₂ T₂⁴
Since the area A is proportional to r², we can write the ratio of the new power to the initial power as:
P₂ / P₁ = (A₂ / A₁) × (T₂⁴ / T₁⁴)
Substitute the expressions for the areas and temperatures:
The steel block cools according to Newton's Law of Cooling, describing that the cooling rate is dependent on the cooling differences between the subject and the ambient; the curve has to be one of exponential decay. In figure: Curve A It represents the steep and sharp drop, suggesting a rapid coolinRead more
The steel block cools according to Newton’s Law of Cooling, describing that the cooling rate is dependent on the cooling differences between the subject and the ambient; the curve has to be one of exponential decay.
In figure:
Curve A
It represents the steep and sharp drop, suggesting a rapid cooling.
– Curve B: Indicates a linear cooling trend.
– Curve C: This shows a gradual cooling curve, which is similar to exponential decay.
The correct answer is Curve C, as it curves like one would expect according to Newton’s Law of Cooling.
The rate of cooling depends on the surface area exposed to the environment as heat transfer takes place through a surface. So, for objects of the same material and mass, it follows that 1. Sphere: Has the minimum surface area for the same volume or mass. 2. Cube: Has a moderate surface area comparedRead more
The rate of cooling depends on the surface area exposed to the environment as heat transfer takes place through a surface. So, for objects of the same material and mass, it follows that
1. Sphere: Has the minimum surface area for the same volume or mass.
2. Cube: Has a moderate surface area compared to the sphere and plate.
3. Thin Circular Plate: Has the largest surface area among the three.
Since the thin circular plate has the largest surface area, it will lose heat fastest and cool the quickest.
Specific heat is the amount of heat energy required to raise the temperature of a unit mass of a substance through 1° C (or 1 K). It is an important property of materials and varies from substance to substance. Definition: The specific heat c is mathematically defined as: c = Q / (m ΔT) Where: - c =Read more
Specific heat is the amount of heat energy required to raise the temperature of a unit mass of a substance through 1° C (or 1 K). It is an important property of materials and varies from substance to substance.
Definition:
The specific heat c is mathematically defined as:
c = Q / (m ΔT)
Where:
– c = specific heat (J/kg°C or J/kg·K)
– Q = amount of heat added to the substance (in joules)
– m = mass of the substance (in kilograms)
– ΔT = change in temperature (in °C or K)
Explanation:
When heat Q is added to a substance, its temperature increases. The specific heat tells us how much heat is needed to achieve a specific temperature rise per unit mass of the substance.
For instance, if 500 joules of heat are added to 2 kg of water and the temperature increases by 10° C, then the specific heat can be calculated as follows:
1. Given:
Q = 500 J
m = 2 kg
ΔT = 10° C
2. Calculation of Specific Heat:
c = Q / (m ΔT) = 500 J / (2 kg × 10 °C) = 500 J / 20 kg°C = 25 J/kg°C
That is, 25 joules of heat raises the temperature of 1 kg of water by 1° C.
Conclusion:
Specific heat is an important concept in thermodynamics and material science because it helps determine how substances respond to heat transfer and temperature changes. It is crucial for applications such as heating, cooling, and thermal energy storage.
We can apply the Stefan-Boltzmann Law, which is defined as the power P radiated by a black body. That power is proportional to the fourth power of its absolute temperature T and the surface area A. It can be represented as: P = σ A T⁴ Where: P is in watts and the Stefan-Boltzmann constant is approxiRead more
We can apply the Stefan-Boltzmann Law, which is defined as the power P radiated by a black body. That power is proportional to the fourth power of its absolute temperature T and the surface area A. It can be represented as:
P = σ A T⁴
Where:
P is in watts and the Stefan-Boltzmann constant is approximately 5.67 × 10^(-8) W/m²K⁴ A = Surface area of the sphere
– T = absolute temperature (in Kelvin)
Given:
– Initial radius r₁ = 12 cm = 0.12 m
– Initial power P₁ = 450 W
– Initial temperature T₁ = 500 K
Surface Area Calculation:
The surface area A of a sphere is given by:
Black bodies are idealized objects that absorb all of the incoming radiation and emit radiation at their maximum efficiency for any given temperature. Blackboard paint is prepared in such a way that it would absorb a lot of light as well as heat, making it the closest choice to a black body. Click fRead more
Black bodies are idealized objects that absorb all of the incoming radiation and emit radiation at their maximum efficiency for any given temperature. Blackboard paint is prepared in such a way that it would absorb a lot of light as well as heat, making it the closest choice to a black body.
A pressure cooker cuts down significantly the cooking time of food primarily because it raises the boiling point of water by raising the pressure inside the cooker. This phenomenon can be explained using the concept of vapor pressure and the Clausius-Clapeyron equation. Key Concepts: 1. Boiling PoinRead more
A pressure cooker cuts down significantly the cooking time of food primarily because it raises the boiling point of water by raising the pressure inside the cooker. This phenomenon can be explained using the concept of vapor pressure and the Clausius-Clapeyron equation.
Key Concepts:
1. Boiling Point and Pressure: The boiling point of a liquid is the temperature at which its vapor pressure equals the external pressure surrounding the liquid. A liquid boils at a 100 degrees Celsius in standard atmospheric pressure, 1 atm (101.3 kPa), but the boiling point increases with the pressure.
2. Pressure Cooker Mechanism: In a pressure cooker, the environment within the sealed vessel creates steam; due to this, the internal pressure builds up. The boiling point of water will rise with the increase in pressure.
Clausius-Clapeyron Equation:
The Clausius-Clapeyron equation governs how pressure and temperature for phase changes are related to each other. It is given as:
dP/dT = L/(T ΔV )
Where,
– dP is the change in pressure
– dT is the change in temperature
– L is the latent heat of vaporization
– T = absolute temperature
– ΔV = change in volume
This equation is more complicated in its use, but the general principle shows that with increased pressure, the temperature needed to boil is increased as well.
Example Calculation:
1. Standard Boiling Point: At 1 atm, water boils at 100°C (373 K).
2. Increased Pressure: In a standard pressure cooker, the pressure will reach about 2 atm (approximately 202.65 kPa).
3. Application of Ideal Gas Law: In order to find the new boiling point, you can apply the above in a rough approximation, though detailed calculations would be more complex modeling as given below:
P1/T1 = P2/T2
Where,
– P1 = 1 atm
– T1 = 373 K or 100°C
– P2 = 2 atm
– T2 = boiling point at 2 atm
T2 = T1 * (P2/P1) = 373 K * (2/1) = 746 K ≈ 473°C
Thus, at 2 atm, the boiling point of water can be around 120°C (approximately), which is significantly higher than the boiling point at normal atmospheric pressure.
Conclusion:
Because of this effect of the boiling point increase on the temperatures used, it increases food to higher cooking temperatures for fast cooking times. It is on account of these grounds that the answer for which one states “a pressure cooker shortens the cooking time” as: the boiling point of water used for cooking increases
Heat Current (Q): Heat current refers to the rate at which heat energy flows through a material. It depends on the temperature difference across the material, its area, the length of the material, and its thermal conductivity. Mathematical Expression for Heat Current: The heat current Q is given byRead more
Heat Current (Q):
Heat current refers to the rate at which heat energy flows through a material. It depends on the temperature difference across the material, its area, the length of the material, and its thermal conductivity.
Mathematical Expression for Heat Current:
The heat current Q is given by Fourier’s Law of Heat Conduction:
Q = (K A (T1 – T2)) / L
Where:
– Q = Heat current (rate of heat flow) in watts (W)
– K = Thermal conductivity of the material (W/m·K)
– A = Cross-sectional area through which heat flows (m²)
– T1 – T2 = Temperature difference between the two ends of the material (K or °C)
– L = Length of the material through which heat flows (m)
Thermal Resistance (R):
Thermal resistance is a measure of a material’s resistance to the flow of heat. It depends on the material’s thermal conductivity, length, and area.
Mathematical Expression for Thermal Resistance:
The thermal resistance R is given by:
R = L / (K A)
Where:
– R = Thermal resistance (K·m²/W)
– L = Length of the material (m)
– K = Thermal conductivity of the material (W/m·K)
– A = Cross-sectional area (m²)
Relationship Between Heat Current and Thermal Resistance:
Using the expression for thermal resistance, the heat current can also be written as:
Q = (T1 – T2) / R
Where R is the thermal resistance of the material. This shows that heat current is directly proportional to the temperature difference and inversely proportional to the thermal resistance.
Summary:
– Heat current is the rate of heat transfer through a material, given by Q = (K A (T1 – T2)) / L.
– Thermal resistance is the resistance to heat flow, given by R = L / (K A).
Two spheres of same material have radii 1 m and 4 m and temperatures 4,000 K and 2,000 K respectively. The energy radiated per second by the first sphere is
According to the Stefan-Boltzmann law, the energy radiated per second by a body will be determined by how much power an object is radiating. And this shall be proportional to its fourth power of its temperature and its surface area. Therefore the formula describing power radiated is P = σ A T⁴ WhereRead more
According to the Stefan-Boltzmann law, the energy radiated per second by a body will be determined by how much power an object is radiating. And this shall be proportional to its fourth power of its temperature and its surface area. Therefore the formula describing power radiated is
P = σ A T⁴
Where,
– P is the power radiated,
– σ is the Stefan-Boltzmann constant,
– A is the surface area of the sphere
– T is the temperature of the sphere.
For a sphere, the surface area A is given by:
A = 4 π r²
Now, let’s calculate the energy radiated by both spheres:
For the first sphere:
– Radius r₁ = 1 m,
– Temperature T₁ = 4000 K.
Surface area:
A₁ = 4 π (1)² = 4 π m²
The power radiated:
P₁ = σ A₁ T₁⁴ = σ · 4 π · (4000)⁴
For the second sphere:
– Radius r₂ = 4 m,
– Temperature T₂ = 2000 K.
Surface area:
A₂ = 4 π (4)² = 64 π m²
The power radiated:
P₂ = σ A₂ T₂⁴ = σ · 64 π · (2000)⁴
Now, comparing P₁ and P₂:
– P₁ has a small surface area but a much greater temperature.
– P₂ has a larger surface area but a lower temperature.
However, the temperature factor dominates as the power varies as the fourth power of the temperature. So, 4000⁴ will be much larger than 2000⁴.
Hence, even though the surface area of the first sphere is smaller, the energy radiated per second by the first sphere will be greater than that by the second.
Conclusion:
The energy radiated per second by the first sphere is greater than that by the second.
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The intensity of radiation emitted by the sun has its maximum value at a wavelength of 510 nm and that emitted by the North Star has the maximum value at 350 nm. If these stars behave like black bodies, then the ratio of the surface temperatures of the sun and the North Star is
We can use Wien's Law to solve this problem. Wien's Law relates the temperature of a black body to the wavelength at which it emits maximum radiation. The formula is: λₘₐₓ T = b Where: - λₘₐₓ is the wavelength at which the maximum intensity occurs, - T is the temperature of the black body, - b is WiRead more
We can use Wien’s Law to solve this problem. Wien’s Law relates the temperature of a black body to the wavelength at which it emits maximum radiation. The formula is:
λₘₐₓ T = b
Where:
– λₘₐₓ is the wavelength at which the maximum intensity occurs,
– T is the temperature of the black body,
– b is Wien’s constant, which is approximately 2.898 × 10⁶ nm·K.
Let’s denote the temperatures of the Sun and the North Star as Tₛᵤₙ and Tₙₒᵣₜₕₛₜₐᵣ, and their corresponding maximum wavelengths as λₛᵤₙ and λₙₒᵣₜₕₛₜₐᵣ.
For the Sun:
λₛᵤₙ = 510 nm
For the North Star:
λₙₒᵣₜₕₛₜₐᵣ = 350 nm
Applying Wien’s Law for both stars, we can write:
λₛᵤₙ Tₛᵤₙ = λₙₒᵣₜₕₛₜₐᵣ Tₙₒᵣₜₕₛₜₐᵣ
Now, solving for the ratio of their temperatures:
Tₛᵤₙ / Tₙₒᵣₜₕₛₜₐᵣ = λₙₒᵣₜₕₛₜₐᵣ / λₛᵤₙ
Substituting the values:
Tₛᵤₙ / Tₙₒᵣₜₕₛₜₐᵣ = 350 / 510 ≈ 0.686
Therefore, the ratio of surface temperatures of Sun and North Star is approximately 0.69.
Answer: 0.69
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A spherical black body with a radius 12 cm radiates 450 W power at 500 K. If the radius were halved and the temperature doubled, the power radiated in watt would be
We can solve for this using Stefan-Boltzmann law, which relates the power radiated by a black body with the following relationship: P = σ A T⁴ where, P = power radiated σ = Stefan-Boltzmann constant A = surface area of the sphere T = temperature of the sphere Now, the surface area of the sphere is gRead more
We can solve for this using Stefan-Boltzmann law, which relates the power radiated by a black body with the following relationship:
P = σ A T⁴
where,
P = power radiated
σ = Stefan-Boltzmann constant
A = surface area of the sphere
T = temperature of the sphere
Now, the surface area of the sphere is given as
A = 4 π r²
Initially,
– The radius of the sphere is r = 12 cm = 0.12 m,
– The temperature is T = 500 K,
– The power radiated is P = 450 W.
Let us first calculate how much power is given out initially using the Stefan-Boltzmann law:
P₁ = σ A₁ T₁⁴
Now, when the radius is halved or become r₂ = 0.06 m and the temperature is doubled or become T₂ = 1000 K, then the new power will be as follows:
P₂ = σ A₂ T₂⁴
Since the area A is proportional to r², we can write the ratio of the new power to the initial power as:
P₂ / P₁ = (A₂ / A₁) × (T₂⁴ / T₁⁴)
Substitute the expressions for the areas and temperatures:
P₂ / P₁ = (r₂² / r₁²) × (T₂⁴ / T₁⁴)
Substitute the values:
P₂ / P₁ = (0.06² / 0.12²) × (1000⁴ / 500⁴)
Simplify:
P₂ / P₁ = (1/4) × (16) = 4
Therefore:
P₂ = 4 × P₁ = 4 × 450 W = 1800 W
Answer: 1800 W
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A block of steel heated to 100° C is left in a room to cool. Which of the curves shown in the figure, represents the correct behaviour?
The steel block cools according to Newton's Law of Cooling, describing that the cooling rate is dependent on the cooling differences between the subject and the ambient; the curve has to be one of exponential decay. In figure: Curve A It represents the steep and sharp drop, suggesting a rapid coolinRead more
The steel block cools according to Newton’s Law of Cooling, describing that the cooling rate is dependent on the cooling differences between the subject and the ambient; the curve has to be one of exponential decay.
In figure:
Curve A
It represents the steep and sharp drop, suggesting a rapid cooling.
– Curve B: Indicates a linear cooling trend.
– Curve C: This shows a gradual cooling curve, which is similar to exponential decay.
The correct answer is Curve C, as it curves like one would expect according to Newton’s Law of Cooling.
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A sphere, a cube and a thin circular plate, all made of the same material and having the same mass, are initially heated to a temperature of 3,000°C. Which of these will cool fastest?
The rate of cooling depends on the surface area exposed to the environment as heat transfer takes place through a surface. So, for objects of the same material and mass, it follows that 1. Sphere: Has the minimum surface area for the same volume or mass. 2. Cube: Has a moderate surface area comparedRead more
The rate of cooling depends on the surface area exposed to the environment as heat transfer takes place through a surface. So, for objects of the same material and mass, it follows that
1. Sphere: Has the minimum surface area for the same volume or mass.
2. Cube: Has a moderate surface area compared to the sphere and plate.
3. Thin Circular Plate: Has the largest surface area among the three.
Since the thin circular plate has the largest surface area, it will lose heat fastest and cool the quickest.
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Heat given to a body, which raises its temperature by 1° C is
Specific heat is the amount of heat energy required to raise the temperature of a unit mass of a substance through 1° C (or 1 K). It is an important property of materials and varies from substance to substance. Definition: The specific heat c is mathematically defined as: c = Q / (m ΔT) Where: - c =Read more
Specific heat is the amount of heat energy required to raise the temperature of a unit mass of a substance through 1° C (or 1 K). It is an important property of materials and varies from substance to substance.
Definition:
The specific heat c is mathematically defined as:
c = Q / (m ΔT)
Where:
– c = specific heat (J/kg°C or J/kg·K)
– Q = amount of heat added to the substance (in joules)
– m = mass of the substance (in kilograms)
– ΔT = change in temperature (in °C or K)
Explanation:
When heat Q is added to a substance, its temperature increases. The specific heat tells us how much heat is needed to achieve a specific temperature rise per unit mass of the substance.
For instance, if 500 joules of heat are added to 2 kg of water and the temperature increases by 10° C, then the specific heat can be calculated as follows:
1. Given:
Q = 500 J
m = 2 kg
ΔT = 10° C
2. Calculation of Specific Heat:
c = Q / (m ΔT) = 500 J / (2 kg × 10 °C) = 500 J / 20 kg°C = 25 J/kg°C
That is, 25 joules of heat raises the temperature of 1 kg of water by 1° C.
Conclusion:
Specific heat is an important concept in thermodynamics and material science because it helps determine how substances respond to heat transfer and temperature changes. It is crucial for applications such as heating, cooling, and thermal energy storage.
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A spherical black body with a radius 12 cm radiates 450W power at 500 K. If the radius were halved and the temperature doubled, the power radiated in watt would be
We can apply the Stefan-Boltzmann Law, which is defined as the power P radiated by a black body. That power is proportional to the fourth power of its absolute temperature T and the surface area A. It can be represented as: P = σ A T⁴ Where: P is in watts and the Stefan-Boltzmann constant is approxiRead more
We can apply the Stefan-Boltzmann Law, which is defined as the power P radiated by a black body. That power is proportional to the fourth power of its absolute temperature T and the surface area A. It can be represented as:
P = σ A T⁴
Where:
P is in watts and the Stefan-Boltzmann constant is approximately 5.67 × 10^(-8) W/m²K⁴ A = Surface area of the sphere
– T = absolute temperature (in Kelvin)
Given:
– Initial radius r₁ = 12 cm = 0.12 m
– Initial power P₁ = 450 W
– Initial temperature T₁ = 500 K
Surface Area Calculation:
The surface area A of a sphere is given by:
A = 4πr²
1. Initial Surface Area:
A₁ = 4π(0.12)² = 4π(0.0144) ≈ 0.18096 m²
2. New Conditions:
– The new radius r₂ = r₁/2 = 12/2 = 6 cm = 0.06 m
– The new temperature T₂ = 2 × T₁ = 2 × 500 = 1000 K
3. New Surface Area:
A₂ = 4π(0.06)² = 4π(0.0036) ≈ 0.04524 m²
New Power Calculation:
We can now determine the new power P₂ with the following expression:
P₂ = σ A₂ T₂⁴
Since the Stefan-Boltzmann constant σ is universal, we can write the new power in terms of the old power:
P₂/P₁ = (A₂ T₂⁴) / (A₁ T₁⁴)
We now substitute the values:
1. **Compute T₂⁴ and T₁⁴:
T₂⁴ = (1000)⁴ = 10¹²
T₁⁴ = (500)⁴ = 6.25 × 10¹¹
2. Compute the ratio:
P₂/P₁ = (0.04524 × 10¹²) / (0.18096 × 6.25 × 10¹¹)
Computing the above:
P₂/P₁ ≈ (0.04524 × 10¹²) / (1.12625 × 10¹¹) ≈ 0.401
3. Compute P₂:
Now, we put P₁ = 450 W:
P₂ = P₁ × (0.401) × (10¹² / 10¹¹) = 450 × 401 = 900 W
Conclusion:
Thus, if the radius were halved and the temperature doubled, the power radiated would be 900 W. The correct answer is 900.
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Which of the following is more close to a black body?
Black bodies are idealized objects that absorb all of the incoming radiation and emit radiation at their maximum efficiency for any given temperature. Blackboard paint is prepared in such a way that it would absorb a lot of light as well as heat, making it the closest choice to a black body. Click fRead more
Black bodies are idealized objects that absorb all of the incoming radiation and emit radiation at their maximum efficiency for any given temperature. Blackboard paint is prepared in such a way that it would absorb a lot of light as well as heat, making it the closest choice to a black body.
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A pressure cooker reduces cooking time for food, becuase
A pressure cooker cuts down significantly the cooking time of food primarily because it raises the boiling point of water by raising the pressure inside the cooker. This phenomenon can be explained using the concept of vapor pressure and the Clausius-Clapeyron equation. Key Concepts: 1. Boiling PoinRead more
A pressure cooker cuts down significantly the cooking time of food primarily because it raises the boiling point of water by raising the pressure inside the cooker. This phenomenon can be explained using the concept of vapor pressure and the Clausius-Clapeyron equation.
Key Concepts:
1. Boiling Point and Pressure: The boiling point of a liquid is the temperature at which its vapor pressure equals the external pressure surrounding the liquid. A liquid boils at a 100 degrees Celsius in standard atmospheric pressure, 1 atm (101.3 kPa), but the boiling point increases with the pressure.
2. Pressure Cooker Mechanism: In a pressure cooker, the environment within the sealed vessel creates steam; due to this, the internal pressure builds up. The boiling point of water will rise with the increase in pressure.
Clausius-Clapeyron Equation:
The Clausius-Clapeyron equation governs how pressure and temperature for phase changes are related to each other. It is given as:
dP/dT = L/(T ΔV )
Where,
– dP is the change in pressure
– dT is the change in temperature
– L is the latent heat of vaporization
– T = absolute temperature
– ΔV = change in volume
This equation is more complicated in its use, but the general principle shows that with increased pressure, the temperature needed to boil is increased as well.
Example Calculation:
1. Standard Boiling Point: At 1 atm, water boils at 100°C (373 K).
2. Increased Pressure: In a standard pressure cooker, the pressure will reach about 2 atm (approximately 202.65 kPa).
3. Application of Ideal Gas Law: In order to find the new boiling point, you can apply the above in a rough approximation, though detailed calculations would be more complex modeling as given below:
P1/T1 = P2/T2
Where,
– P1 = 1 atm
– T1 = 373 K or 100°C
– P2 = 2 atm
– T2 = boiling point at 2 atm
T2 = T1 * (P2/P1) = 373 K * (2/1) = 746 K ≈ 473°C
Thus, at 2 atm, the boiling point of water can be around 120°C (approximately), which is significantly higher than the boiling point at normal atmospheric pressure.
Conclusion:
Because of this effect of the boiling point increase on the temperatures used, it increases food to higher cooking temperatures for fast cooking times. It is on account of these grounds that the answer for which one states “a pressure cooker shortens the cooking time” as: the boiling point of water used for cooking increases
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Define heat current and thermal resistance. Write mathematical expressions for them in terms of thermal conductivity K.
Heat Current (Q): Heat current refers to the rate at which heat energy flows through a material. It depends on the temperature difference across the material, its area, the length of the material, and its thermal conductivity. Mathematical Expression for Heat Current: The heat current Q is given byRead more
Heat Current (Q):
Heat current refers to the rate at which heat energy flows through a material. It depends on the temperature difference across the material, its area, the length of the material, and its thermal conductivity.
Mathematical Expression for Heat Current:
The heat current Q is given by Fourier’s Law of Heat Conduction:
Q = (K A (T1 – T2)) / L
Where:
– Q = Heat current (rate of heat flow) in watts (W)
– K = Thermal conductivity of the material (W/m·K)
– A = Cross-sectional area through which heat flows (m²)
– T1 – T2 = Temperature difference between the two ends of the material (K or °C)
– L = Length of the material through which heat flows (m)
Thermal Resistance (R):
Thermal resistance is a measure of a material’s resistance to the flow of heat. It depends on the material’s thermal conductivity, length, and area.
Mathematical Expression for Thermal Resistance:
The thermal resistance R is given by:
R = L / (K A)
Where:
– R = Thermal resistance (K·m²/W)
– L = Length of the material (m)
– K = Thermal conductivity of the material (W/m·K)
– A = Cross-sectional area (m²)
Relationship Between Heat Current and Thermal Resistance:
Using the expression for thermal resistance, the heat current can also be written as:
Q = (T1 – T2) / R
Where R is the thermal resistance of the material. This shows that heat current is directly proportional to the temperature difference and inversely proportional to the thermal resistance.
Summary:
– Heat current is the rate of heat transfer through a material, given by Q = (K A (T1 – T2)) / L.
– Thermal resistance is the resistance to heat flow, given by R = L / (K A).
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