When the 3.0 kg cylinder fell 500 m, the final temperature of the water was °C and the change in temperature was °C.

When the 9.0 kg cylinder fell 500 m, the final temperature of the water was °C and the change in temperature was °C.

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