Answer:
The change in temperature of the water due to the first cylinder, ΔT = 0.82 ° C
The final temperature of the water, T₀ = 20.82° C
The change in temperature due to the second cylinder, ΔT' = 2.45° C
The final temperature of the water, T' = 23.27° C
Explanation:
Given data,
The mass of the cylinder, m = 3.0 kg
The cylinder fell from the height, h = 500 m
The mass of the second cylinder, M = 9 kg
Te second cylinder fell from the height, h' = 500 m
The normal temperature of the water is, T = 20° C
The mass of the water, m' = 10 kg
During a collision with the water, consider the kinetic energy is entirely converted into heat energy.
The K.E of the first cylinder falling from height h to the surface of the water is equal to the P.E at that height.
K.E = P.E
P.E = mgh
= 3 x 9.8 x 500
= 14700 J
The change in temperature,
ΔT = E / m c
= 14700 / (10 x 1800)
= 0.82° C
The change in temperature of the water due to the first cylinder, ΔT = 0.82 ° C
The final temperature of water,
T₀ = T + ΔT
= 20° C + 0.82° C
= 20.82° C
The final temperature of the water, T₀ = 20.82° C
The K.E of the second cylinder falling from height h to the surface of the water is,
P.E = Mgh'
= 9 x 9.8 x 500
= 44100 J
The change in temperature,
ΔT' = E' / m c
= 44100 / (10 x 1800)
= 2.45° C
The change in temperature due to the second cylinder, ΔT' = 2.45° C
The final temperature of water,
T' = T + ΔT + ΔT'
= 20° C + 0.82 ° C + 2.45° C
= 23.27° C
Hence, the final temperature of the water, T' = 23.27° C
