Water at Bhakra dam falls through a height of 210 m. Assuming that the whole of the energy due to fall is converted into heat, calculate the rise in temperature of water. Take $J = 4.2Jca{l^{ - 1}}$
and $g = 9.8m{s^{ - 2}}$.
Answer
603k+ views
Hint: There is one law which will be valid anywhere and at any situation universally and that law is law of conservation of energy. That means energy can neither be created nor destroyed but it is only converted from one form to the other. By using this law we will solve this problem.
Formula used:
$\eqalign{
& U = mgh \cr
& Q = mc\Delta T \cr} $
Complete step by step answer:
It is given that water is falling from the dam to the ground. Now when the motion of the body is along the direction of conservative force then potential energy is lost and if the motion of the body is against the conservative force then the potential energy is gained.
Here in this problem water is falling from the dam to the ground which means that water is moving along the conservative force which is the gravitational force over here.
So it is clear that when it happens potential energy would be lost and it is also given that all the lost energy is utilized to heat the water.
Now when the heat is given to the water obviously the temperature of the water goes up.
The potential energy lost will be $U = mgh$
Where ‘m’ is the mass of the water and ‘g’ is the acceleration due to gravity and ‘h’ is the height difference between dam and the ground.
The heat consumed by the water will be $Q = mc\Delta T$
Where ‘m’ is the mass and ‘c’ is the specific heat of water and ‘$\Delta T$’ is the temperature raise.
So we equate the potential energy loss with the heat gained
$U = Q$
$\eqalign{
& \Rightarrow mgh = mc\Delta T \cr
& \Rightarrow gh = c\Delta T \cr
& \Rightarrow \Delta T = \dfrac{{gh}}{c} \cr
& \Rightarrow \Delta T = \dfrac{{9.8 \times 210}}{{4.2 \times {{10}^3}}} \cr
& \Rightarrow \Delta T = 0.49K \cr} $
So the temperature raise would be 0.49 kelvin.
Note:
Law of conservation of energy is valid everywhere but the law of conservation of energy is valid only when no non conservative force acts. Here kinetic energy of the water is assumed to be constant. In the final temperature raise expression we can express the same in Celsius too.
Formula used:
$\eqalign{
& U = mgh \cr
& Q = mc\Delta T \cr} $
Complete step by step answer:
It is given that water is falling from the dam to the ground. Now when the motion of the body is along the direction of conservative force then potential energy is lost and if the motion of the body is against the conservative force then the potential energy is gained.
Here in this problem water is falling from the dam to the ground which means that water is moving along the conservative force which is the gravitational force over here.
So it is clear that when it happens potential energy would be lost and it is also given that all the lost energy is utilized to heat the water.
Now when the heat is given to the water obviously the temperature of the water goes up.
The potential energy lost will be $U = mgh$
Where ‘m’ is the mass of the water and ‘g’ is the acceleration due to gravity and ‘h’ is the height difference between dam and the ground.
The heat consumed by the water will be $Q = mc\Delta T$
Where ‘m’ is the mass and ‘c’ is the specific heat of water and ‘$\Delta T$’ is the temperature raise.
So we equate the potential energy loss with the heat gained
$U = Q$
$\eqalign{
& \Rightarrow mgh = mc\Delta T \cr
& \Rightarrow gh = c\Delta T \cr
& \Rightarrow \Delta T = \dfrac{{gh}}{c} \cr
& \Rightarrow \Delta T = \dfrac{{9.8 \times 210}}{{4.2 \times {{10}^3}}} \cr
& \Rightarrow \Delta T = 0.49K \cr} $
So the temperature raise would be 0.49 kelvin.
Note:
Law of conservation of energy is valid everywhere but the law of conservation of energy is valid only when no non conservative force acts. Here kinetic energy of the water is assumed to be constant. In the final temperature raise expression we can express the same in Celsius too.
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