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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?

Sagot :

Answer:

m_added = 2 kg

Explanation:

From the question, we are told that the cylinder is allowed to fall 800 m in height. Thus, the potential energy will be converted into heat energy which will increase the temperature of water .

Now, let the mass of the falling cylinder be denoted by "m1" and let h be the height of fall.

Thus;

Formula for potential energy = mgh

Thus, as said earlier it's converted to heat generated. So heat generated = m1gh

Now let's calculate the heat absorbed;

heat absorbed = (m2)cΔt

Where;

ΔT is change in temperature

c is specific heat of water .

m2 is mass of water

Heat absorbed = heat generated

Thus;

(m2)cΔt = m1gh

Δt = m1gh/(m2•c)

Now, in both cases of the water and cylinder, m1, g , h and c are constant

Thus, we have;

Δt = (m1gh/m2) × 1/c

Where;

(m1gh/m2) is denoted as a constant k.

Thus;

Δt = K/m

For the first experiment, we have;

m = 2 kg

Δt = 2.4

Thus;

2.4 = K/2

Multiply both sides by 2 to get;

K = 4.8

For the second experiment, we have;

Δt = 1.2

Also,we have seen that K = 4.8

Thus;

Δt = K/m

Thus;

1.2 = 4.8/m

m = 1.2

m = 4 kg

Thus,mass added is;

m_added = 4 - 2

m_added = 2 kg

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