The device shown below contains 2 kg of water. The cylinder is allowed to fall 800 m during which the temperature of the water increases by 2.4°C. Some amount of water is added to the container and the experiment is repeated. All other values remain constant. This time the temperature of the water increases by 1.2°C. How much water was added to the container?

A. 0.8 kg
B. 1 kg
C. 2 kg
D. 3.6 kg

C. 2 kg

Explanation:

From the problem , it appears that due to fall by 800m (h )  , potential energy is converted into heat energy which increases the temperature of water . Let the mass of cylinder falling be M , h be the height

loss of potential energy = Mgh

heat generated = Mgh

let heat generated increases a temperature rise of Δt in the mass of water m

heat absorbed = msΔt , s is specific heat of water .

msΔt = Mgh

Δt = Mgh / ms

Δt = K / m ( M , g , h and s are constant in both the case)

For first case  ,

m = 2 kg and Δt  = 2.4

2.4 = K / 2

K = 4.8

For second case

Δt = K / m

1.2 = 4.8 / m

m = 4 kg

mass added = 4 - 2

= 2 kg .