Question 178: Thermodynamics & KTG

Question

Water falls from a height of 200 m into a pool. Calculate the rise in temperature of the water assuming no heat dissipation from the water in the pool.(Take g = 10 m/s2, specific heat of water = 4200 J/(kg K))

Accepted Answer

To find the rise in temperature of the water as it falls from a height, we start by equating the potential energy loss to the heat gained, assuming no heat is lost from the water in the pool.
The potential energy lost by the water can be expressed as:
$ mgh $
where:
$ m $ is the mass of water,
$ g $ is the acceleration due to gravity (10 m/sÃÂ²),
$ h $ is the height (200 m).
The heat gained by the water when temperature changes is given by:
$ ms \Delta T $
where:
$ s $ is the specific heat capacity of water (4200 J/(kg K)),
$ \Delta T $ is the change in temperature.
Setting the potential energy loss equal to the heat gained:
$ mgh = ms \Delta T $
We can simplify this equation by canceling out the mass $ m $:
$ gh = s \Delta T $
Now, solve for $ \Delta T $:
$ \Delta T = \frac{gh}{s} $
Substitute the known values:
$ \Delta T = \frac{10 	imes 200}{4200} $
Calculate $ \Delta T $:
$ \Delta T = \frac{2000}{4200} $
Simplify the fraction:
$ \Delta T = \frac{10}{21} $
Therefore, the rise in temperature is $ \Delta T \approx 0.48 \, 	ext{K} $.

Answer

0.36 K

Answer

0.23 K

Answer

0.48 K

Answer

0.14 K

Thermodynamics & KTG

Water falls from a height of 200 m into a pool. Calculate the rise in temperature of the water assuming no heat dissipation from the water in the pool.(Take g = 10 m/s2, specific heat of water = 4200 J/(kg K))

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