Let's analyze the energy transfer to solve this problem. Step 1: Energy released by X g of steam When steam at 100°C condenses to water at 100°C, the energy released is: Qₛₜₑₐₘ = X ⋅ Lᵥ where Lᵥ = 540 cal/g (latent heat of vaporization of water). Step 2: Energy absorbed by Y g of ice The ice first mRead more
Let’s analyze the energy transfer to solve this problem.
Step 1: Energy released by X g of steam
When steam at 100°C condenses to water at 100°C, the energy released is:
Qₛₜₑₐₘ = X ⋅ Lᵥ
where Lᵥ = 540 cal/g (latent heat of vaporization of water).
Step 2: Energy absorbed by Y g of ice
The ice first melts into water at 0°C, and then this water is heated to 100°C. The total energy absorbed by Y g of ice is:
Qᵢcₑ = Y ⋅ Lf + Y ⋅ c ⋅ ΔT
where:
– Lf = 80 cal/g (latent heat of fusion of ice),
– c = 1 cal/g°C (specific heat capacity of water),
– ΔT = 100°C.
Put values:
Qᵢcₑ = Y ⋅ 80 + Y ⋅ 1 ⋅ 100 = Y ⋅ (80 + 100) = Y ⋅ 180 cal.
Step 3: Energy conservation
The energy released by steam is equal to the energy absorbed by ice:
X ⋅ 540 = Y ⋅ 180
To solve this we analyze this situation by applying the concept of energy conversion as follows; Step 1: Potential Energy converted to heat To indicate that whenever water drops from a height of 500 m, its potential energy is turned into heat. The PE per unit mass is thus given as: PE = mgh where: -Read more
To solve this we analyze this situation by applying the concept of energy conversion as follows;
Step 1: Potential Energy converted to heat To indicate that whenever water drops from a height of 500 m, its potential energy is turned into heat. The PE per unit mass is thus given as: PE = mgh where: – m is the mass of water in kilograms
– g = 9.8 m/s² is the acceleration due to gravity,
– h = 500 m is the height.
So:
PE = m ⋅ 9.8 ⋅ 500
Step 2: Energy to temperature rise
The heat produced is utilized to raise the temperature of water. The heat equation is:
Q = m ⋅ c ⋅ ΔT
where:
– Q is the heat energy,
– c = 4300 J/kg°C is the specific heat of water,
– ΔT is the rise in temperature.
Q = PE:
m ⋅ 9.8 ⋅ 500 = m ⋅ 4300 ⋅ ΔT
Step 3: Simplify and solve for ΔT
Cancel m from both sides:
9.8 ⋅ 500 = 4300 ⋅ ΔT
When a force acts on a body, and its line of action does not pass through the center of gravity, the body experiences two effects: linear acceleration and angular acceleration. Linear acceleration occurs as the force pushes or pulls the body in a specific direction. This is due to the body's responsRead more
When a force acts on a body, and its line of action does not pass through the center of gravity, the body experiences two effects: linear acceleration and angular acceleration. Linear acceleration occurs as the force pushes or pulls the body in a specific direction. This is due to the body’s response to the applied force, which causes a change in its state of motion.
At the same time, since the line of action of this force does not pass through the center of gravity, a torque is developed. Torque is defined as the rotational effect developed due to the application of the force at a distance from the center of gravity. Thus, this torque causes angular acceleration, and hence, the body rotates around an axis. The combined effects of the linear acceleration and angular acceleration influence the motion of the body.
For example, when you push a door at its edge, it rotates around its hinges because of the torque generated, and the force exerts a linear effect as well. Similarly, if you apply force to a spinning top at any point other than its center, it causes rotation and a change in position as well. So, whenever the line of action of the force misses the center of gravity, both types of acceleration occur.
There was said to work when a force is applied onto an object along with the movements of the applied force. But for work, however to be done then three conditions apply: 1. Applied Force: One must apply forces on the affected object. 2. Displacement by the Applied force: The resultant movement of aRead more
There was said to work when a force is applied onto an object along with the movements of the applied force. But for work, however to be done then three conditions apply:
1. Applied Force: One must apply forces on the affected object.
2. Displacement by the Applied force: The resultant movement of applying the force with the object having moved from position.
3. Direction Alignment: The displacement must have a component in the direction of the applied force.
Examples of Work:
1. Lifting an Object: When you lift a book from the ground, you apply an upward force, and the book moves upwards. Here, work is done because the force and displacement are in the same direction.
2. Pushing a cart: While the person applies the force for pushing the shopping cart, work gets done due to the displacement in the direction of the force.
3. Pulling a Sled: One example of using force to do work would be pulling the sled on snowy surfaces. Applying force to push it in the direction of pull means work done is achieved.
4. Lifting Water Using a Bucket: If a bucket is used to draw water from a well, then work is done as the bucket goes upwards because of the applied force.
Work is not done when the object is stationary or the applied force is perpendicular to the displacement. This means holding an immovable object or carrying a load horizontally in which no vertical displacement is created does not include work.
When two bodies of masses m and 4m have the same amount of kinetic energy, their momenta differ because the relationship between kinetic energy and momentum is such that kinetic energy depends on both mass and the square of velocity, while momentum depends linearly on mass and velocity. The velocityRead more
When two bodies of masses m and 4m have the same amount of kinetic energy, their momenta differ because the relationship between kinetic energy and momentum is such that kinetic energy depends on both mass and the square of velocity, while momentum depends linearly on mass and velocity.
The velocity of a heavier body will have to be lower than that of a lighter body in order to have the same kinetic energy. For the kinetic energy being constant, the momentum of a body varies directly as the square root of its mass. So when their momenta are compared, the ratio of the momenta is equal to the square root of the ratio of the masses.
In this case, the first body has mass m, while the second has a mass of 4m. The square root of their mass ratio, √1} : √4, gives the momentum ratio as 1:2. This means the body with four times the mass has double the momentum of the lighter body under equal kinetic energy conditions.
Hence, the kinetic energy of both bodies is the same, but their momentum differs because of the mass in these bodies. It is greater in the former body compared with the later.
In an energy recycling process, X g of steam at 100 °C becomes water at 100 °C which converts Y g of ice at 0 °C into water at 100 °C. The ratio of X and Y will be
Let's analyze the energy transfer to solve this problem. Step 1: Energy released by X g of steam When steam at 100°C condenses to water at 100°C, the energy released is: Qₛₜₑₐₘ = X ⋅ Lᵥ where Lᵥ = 540 cal/g (latent heat of vaporization of water). Step 2: Energy absorbed by Y g of ice The ice first mRead more
Let’s analyze the energy transfer to solve this problem.
Step 1: Energy released by X g of steam
When steam at 100°C condenses to water at 100°C, the energy released is:
Qₛₜₑₐₘ = X ⋅ Lᵥ
where Lᵥ = 540 cal/g (latent heat of vaporization of water).
Step 2: Energy absorbed by Y g of ice
The ice first melts into water at 0°C, and then this water is heated to 100°C. The total energy absorbed by Y g of ice is:
Qᵢcₑ = Y ⋅ Lf + Y ⋅ c ⋅ ΔT
where:
– Lf = 80 cal/g (latent heat of fusion of ice),
– c = 1 cal/g°C (specific heat capacity of water),
– ΔT = 100°C.
Put values:
Qᵢcₑ = Y ⋅ 80 + Y ⋅ 1 ⋅ 100 = Y ⋅ (80 + 100) = Y ⋅ 180 cal.
Step 3: Energy conservation
The energy released by steam is equal to the energy absorbed by ice:
X ⋅ 540 = Y ⋅ 180
Step 4: Solve for X/Y
X / Y = 180 / 540 = 1 / 3
Final Answer:
The ratio of X and Y is 1:3.
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Water falls from a height 500 m. What is the rise in the temperature of water at the bottom, if the whole energy remains in water?(Specific heat of water = 4300 J/kg °C)
To solve this we analyze this situation by applying the concept of energy conversion as follows; Step 1: Potential Energy converted to heat To indicate that whenever water drops from a height of 500 m, its potential energy is turned into heat. The PE per unit mass is thus given as: PE = mgh where: -Read more
To solve this we analyze this situation by applying the concept of energy conversion as follows;
Step 1: Potential Energy converted to heat To indicate that whenever water drops from a height of 500 m, its potential energy is turned into heat. The PE per unit mass is thus given as: PE = mgh where: – m is the mass of water in kilograms
– g = 9.8 m/s² is the acceleration due to gravity,
– h = 500 m is the height.
So:
PE = m ⋅ 9.8 ⋅ 500
Step 2: Energy to temperature rise
The heat produced is utilized to raise the temperature of water. The heat equation is:
Q = m ⋅ c ⋅ ΔT
where:
– Q is the heat energy,
– c = 4300 J/kg°C is the specific heat of water,
– ΔT is the rise in temperature.
Q = PE:
m ⋅ 9.8 ⋅ 500 = m ⋅ 4300 ⋅ ΔT
Step 3: Simplify and solve for ΔT
Cancel m from both sides:
9.8 ⋅ 500 = 4300 ⋅ ΔT
Solve for ΔT: ΔT = (9.8 ⋅ 500) / 4300 = 4900 / 4300 ≈ 1.16°C
Final Answer:
The temperature of the water at the bottom has increased by 1.16°C.
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See lesshttps://www.tiwariacademy.com/ncert-solutions/class-11/physics/chapter-10/
If force acts on a body, whose line of action does not pass through its centre of gravity, then the body will experience
When a force acts on a body, and its line of action does not pass through the center of gravity, the body experiences two effects: linear acceleration and angular acceleration. Linear acceleration occurs as the force pushes or pulls the body in a specific direction. This is due to the body's responsRead more
When a force acts on a body, and its line of action does not pass through the center of gravity, the body experiences two effects: linear acceleration and angular acceleration. Linear acceleration occurs as the force pushes or pulls the body in a specific direction. This is due to the body’s response to the applied force, which causes a change in its state of motion.
At the same time, since the line of action of this force does not pass through the center of gravity, a torque is developed. Torque is defined as the rotational effect developed due to the application of the force at a distance from the center of gravity. Thus, this torque causes angular acceleration, and hence, the body rotates around an axis. The combined effects of the linear acceleration and angular acceleration influence the motion of the body.
For example, when you push a door at its edge, it rotates around its hinges because of the torque generated, and the force exerts a linear effect as well. Similarly, if you apply force to a spinning top at any point other than its center, it causes rotation and a change in position as well. So, whenever the line of action of the force misses the center of gravity, both types of acceleration occur.
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See lesshttps://www.tiwariacademy.com/ncert-solutions/class-11/physics/chapter-5/
When is the work said to be done? Give some examples.
There was said to work when a force is applied onto an object along with the movements of the applied force. But for work, however to be done then three conditions apply: 1. Applied Force: One must apply forces on the affected object. 2. Displacement by the Applied force: The resultant movement of aRead more
There was said to work when a force is applied onto an object along with the movements of the applied force. But for work, however to be done then three conditions apply:
1. Applied Force: One must apply forces on the affected object.
2. Displacement by the Applied force: The resultant movement of applying the force with the object having moved from position.
3. Direction Alignment: The displacement must have a component in the direction of the applied force.
Examples of Work:
1. Lifting an Object: When you lift a book from the ground, you apply an upward force, and the book moves upwards. Here, work is done because the force and displacement are in the same direction.
2. Pushing a cart: While the person applies the force for pushing the shopping cart, work gets done due to the displacement in the direction of the force.
3. Pulling a Sled: One example of using force to do work would be pulling the sled on snowy surfaces. Applying force to push it in the direction of pull means work done is achieved.
4. Lifting Water Using a Bucket: If a bucket is used to draw water from a well, then work is done as the bucket goes upwards because of the applied force.
Work is not done when the object is stationary or the applied force is perpendicular to the displacement. This means holding an immovable object or carrying a load horizontally in which no vertical displacement is created does not include work.
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See lesshttps://www.tiwariacademy.com/ncert-solutions/class-11/physics/chapter-5/
Two bodies of mass m and 4 m have equal kinetic energy. What is the ratio of their momentum?
When two bodies of masses m and 4m have the same amount of kinetic energy, their momenta differ because the relationship between kinetic energy and momentum is such that kinetic energy depends on both mass and the square of velocity, while momentum depends linearly on mass and velocity. The velocityRead more
When two bodies of masses m and 4m have the same amount of kinetic energy, their momenta differ because the relationship between kinetic energy and momentum is such that kinetic energy depends on both mass and the square of velocity, while momentum depends linearly on mass and velocity.
The velocity of a heavier body will have to be lower than that of a lighter body in order to have the same kinetic energy. For the kinetic energy being constant, the momentum of a body varies directly as the square root of its mass. So when their momenta are compared, the ratio of the momenta is equal to the square root of the ratio of the masses.
In this case, the first body has mass m, while the second has a mass of 4m. The square root of their mass ratio, √1} : √4, gives the momentum ratio as 1:2. This means the body with four times the mass has double the momentum of the lighter body under equal kinetic energy conditions.
Hence, the kinetic energy of both bodies is the same, but their momentum differs because of the mass in these bodies. It is greater in the former body compared with the later.
Click here :
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-11/physics/chapter-5/