
Water falls from a height 420 m. What is the rise in its temperature?
Answer
577.2k+ views
Hint
We are assuming that all of its potential energy is converted into heat. Potential energy is energy that is stored – or conserved - in an object or substance. This stored energy is based on the position, arrangement or state of the object or substance. You can think of it as energy that has the 'potential' to do work.
Complete step by step answer
According to the question , Water is at height (h)=420m.
We know that, Potential energy $E = mgh$
Here, g = gravitational force = 9.81 m/ s2 and h = 420 m
$E = 1 \times 9.81 \times 420$
$E = 4120.2$ J/litre
We know, Specific heat of water = 4180 J/kg k.
Now, as we assume whole potential energy converted into heat ,
So, $E = mc\Delta T$ ...............[c = specific heat]
$ \Rightarrow 4120.2 = 1 \times 4180 \times \Delta T$
$ \Rightarrow \Delta T = \dfrac{{4120.2}}{{4180}}$
$\therefore \Delta T = 0.98 = {1^\circ }C$ (approximately).
Note
Specific heat is a particular type of heat capacity. Specific heat is the thermodynamic property, which states the amount of heat required for a single unit of mass of a substance to be raised by one degree of temperature. In physics, the law of conservation of energy states that the total energy of an isolated system remains constant; it is said to be conserved over time.
We are assuming that all of its potential energy is converted into heat. Potential energy is energy that is stored – or conserved - in an object or substance. This stored energy is based on the position, arrangement or state of the object or substance. You can think of it as energy that has the 'potential' to do work.
Complete step by step answer
According to the question , Water is at height (h)=420m.
We know that, Potential energy $E = mgh$
Here, g = gravitational force = 9.81 m/ s2 and h = 420 m
$E = 1 \times 9.81 \times 420$
$E = 4120.2$ J/litre
We know, Specific heat of water = 4180 J/kg k.
Now, as we assume whole potential energy converted into heat ,
So, $E = mc\Delta T$ ...............[c = specific heat]
$ \Rightarrow 4120.2 = 1 \times 4180 \times \Delta T$
$ \Rightarrow \Delta T = \dfrac{{4120.2}}{{4180}}$
$\therefore \Delta T = 0.98 = {1^\circ }C$ (approximately).
Note
Specific heat is a particular type of heat capacity. Specific heat is the thermodynamic property, which states the amount of heat required for a single unit of mass of a substance to be raised by one degree of temperature. In physics, the law of conservation of energy states that the total energy of an isolated system remains constant; it is said to be conserved over time.
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