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The water at Niagara Falls drops through a height of 55.0 m. If the water’s loss of gravitational potential energy shows up as an increase in temperature of the water, what is the temperature difference between the water at the top of the falls and the water at the bottom? For this problem, if you need any of these values, use g = 10 m/s^2, take the specific heat of water to be 4000 J/(kg °C), and the latent heat of vaporization of water to be 2 × 10^6 J/kg.__________.

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Answer:

ΔT = 0.14 °C

Step-by-step explanation:

The gravitational potential energy of the water can be calculated using the following expression.

Ep = m . g . h

where,

m: mass

g: gravity

h: height

The heat used to increase the temperature of the water can be calculated using the following expression.

Q = c . m . ΔT

where,

c: specific heat capacity

m: mass

ΔT: difference in temperature

According to the law of conservation of energy, the energy lost from gravitational potential energy is equal to the heat gained by the water. Then,

Ep = Q

m . g . h = c . m . ΔT

g . h = c . ΔT

(10 m/s²) × (55.0 m) = (4000 J/kg°C) × ΔT

ΔT = 0.14 °C

User GregB
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