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 $g = 10 \, \text{m/s}^2$, specific heat of water = $4200 \, \text{J/(kg K)}$)
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Solution Steps
1
Identify Energy Conversion
As water falls from height $h$, its gravitational potential energy (PE) is converted into thermal energy ($Q$) upon impact with the pool.
Since there is no heat dissipation, $\Delta PE = Q$.
2
Set up the Equation
Potential Energy, $PE = mgh$
Heat absorbed by water, $Q = ms\Delta T$
Equating both: $mgh = ms\Delta T$
3
Simplify for Temperature Rise ($\Delta T$)
The mass $m$ cancels out from both sides, indicating the temperature rise is independent of the amount of water falling:
$$gh = s\Delta T \implies \Delta T = \frac{gh}{s}$$
4
Substitute the Given Values
Height $h = 200 \, \text{m}$, $g = 10 \, \text{m/s}^2$, and specific heat $s = 4200 \, \text{J/kg K}$.
$$\Delta T = \frac{10 \times 200}{4200}$$
$$\Delta T = \frac{2000}{4200} = \frac{20}{42}$$
5
Final Calculation
$$\Delta T = \frac{10}{21} \approx 0.47619… \, \text{K}$$
Rounding to two decimal places as per options:
$\Delta T = 0.48 \, \text{K}$
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Theory
1. Conservation of Mechanical Energy
In the context of falling fluids, mechanical energy in the form of potential energy ($mgh$) is initially held at the top. As the fluid falls, it gains kinetic energy. When it hits the surface of the pool, the sudden deceleration converts this kinetic energy into internal molecular motion, which we observe as heat. This is an application of the First Law of Thermodynamics, where work done by gravity is transformed into internal energy ($\Delta U$).
2. Specific Heat Capacity ($s$)
Specific heat capacity is the amount of heat energy required to raise the temperature of a unit mass of a substance by one degree Celsius (or one Kelvin). For water, the value is remarkably high ($\approx 4200 \, \text{J/kg K}$), which means water can absorb a significant amount of energy with relatively small changes in temperature. This explains why the temperature rise in the problem is quite small (less than $0.5 \, \text{K}$) despite a significant fall of 200 meters.
3. Mechanical Equivalent of Heat
Historically, James Prescott Joule performed experiments similar to this falling-water scenario to determine the mechanical equivalent of heat. He used falling weights to turn a paddle in water, proving that mechanical work and heat are interchangeable forms of energy. In modern SI units, the conversion factor is $1$, since both work and heat are measured in Joules ($J$). This problem assumes an ideal scenario where 100% of the potential energy is utilized to heat the water.
4. Factors in Practical Scenarios
In a real-world waterfall, several factors prevent the temperature rise from being exactly $\Delta T = gh/s$. These include air resistance (which dissipates some energy as heat into the air before the water hits the pool), evaporation (which cools the water), and sound energy produced upon impact. Furthermore, if the pool is flowing or large, the heat dissipates into the surroundings. However, for JEE problems, we follow the “no dissipation” constraint to simplify the calculation.
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FAQs
1
Does the volume of the pool affect the result?
In this specific calculation, we are finding the rise in temperature of the “falling water” itself upon impact. If it mixes with a much larger pool, the overall temperature rise of the pool would be much smaller.
2
Is the answer the same in Celsius and Kelvin?
Yes, because we are calculating a change in temperature ($\Delta T$), and the magnitude of 1 degree Celsius is equal to 1 Kelvin.
3
What if air resistance was not negligible?
The temperature rise would be smaller because some energy would be lost to the air during the fall.
4
Why did mass cancel out?
Because both the potential energy available and the energy required for a temperature change are directly proportional to mass.
5
Can this method be used for other liquids?
Yes, you would simply replace the specific heat capacity $s$ with that of the liquid (e.g., oil or mercury).
6
Is 200 m a realistic height for this effect?
Yes, Victoria Falls is about 108 m and Angel Falls is 979 m. In very high falls, a measurable temperature difference can indeed be observed between the top and bottom.
7
Does the density of water change the answer?
No, density affects volume but not the mass-energy relationship used in $mgh = ms\Delta T$.
8
What if the water was already boiling?
If the water was at $100^\circ\text{C}$, the energy might go into latent heat of vaporization (turning some water into steam) instead of increasing temperature.
9
Is the velocity of the water at impact relevant?
Velocity $v = \sqrt{2gh}$ is the intermediate state of energy (Kinetic Energy). We bypass it by using the total energy conversion from Potential to Thermal.
10
What is the significance of the “no heat dissipation” assumption?
It means we are treating the system as adiabatic, where no energy leaves the water to the air or the ground.
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