Question

James Joule (after whom the unit of energy is named) claimed that the water at the bottom of Niagara Falls should be warmer than the water at the top, 51 m above the bottom. He reasoned that the falling water would transform its gravitational potential energy at the top into thermal energy at the bottom, where turbulence brings the water almost to a halt.. If this transformation is the only process occurring, how much warmer will the water at the bottom be? The specific heat of water is 4190 J/(kg?K).

Answer 1

The temperature increase at the bottom of Niagara Falls due to gravitational potential energy conversion into thermal energy is approximately 0.119 degrees Celsius for a 51-meter fall, calculated by the heat gained equating to the lost potential energy.

Step-by-step explanation:

To calculate the temperature increase of the water at the bottom of Niagara Falls, we use the principle that gravitational potential energy is converted into thermal energy. When water at a higher elevation falls, it loses gravitational potential energy, which is given by the formula Potential Energy (PE) = mgh, where m is the mass, g is the acceleration due to gravity (9.8 m/s2), and h is the height of the fall.

Once the water is at the bottom, the lost potential energy will be equal to the heat energy gained due to the principle of conservation of energy. This gain in heat energy can cause an increase in the water's temperature, calculated using the formula Q = mcΔT, where Q is the heat energy, m is the mass, c is the specific heat of water, and ΔT is the change in temperature.

Setting these two energy equations equal to each other, we solve for ΔT as follows:

mgh = mcΔT

ΔT = gh/c

For a 51-meter fall and with the specific heat c of water being 4190 J/(kg·K),

ΔT = (9.8 m/s2)(51 m) / 4190 J/(kg·K) = 0.119 K or °C

Thus, the water's temperature increase will be approximately 0.119 degrees Celsius after falling 51 meters, assuming no other heat losses or gains.

Answer 2

0.12 K

Step-by-step explanation:

height, h = 51 m

let the mass of water is m.

Specific heat of water, c = 4190 J/kg K

According to the transformation of energy

Potential energy of water = thermal energy of water

m x g x h = m x c x ΔT

Where, ΔT is the rise in temperature

g x h = c x ΔT

9.8 x 51 = 4190 x ΔT

ΔT = 0.12 K

Thus, the rise in temperature is 0.12 K.