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

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