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.