Question

In a device like the one shown below, the cylinder is allowed to fall a distance of 300 m. As a result, the temperature of the water increases by 2.7°C. What would have been the increase in temperature if the cylinder were only allowed to fall 100 m?

Answer 1

Answer:

Increase in the temperature of water would be 0.9 degree C

Explanation:

As we know by energy conservation

Change in the gravitational potential energy of the cylinder = increase in the thermal energy of the water

Here we know that the gravitational potential energy of the cylinder is given as

$U = mgh$

here we have

h = 300 m

now we can say

$Mc\Delta T = (m \times 9.8 \times 300)$

now if the cylinder falls from height h = 100 m

then we have

$Mc\Delta T' = (m \times 9.8 \times 100)$

now from above two equations

$\frac{\Delta T'}{\Delta T} = \frac{100}{300}$

$\Delta T' = 2.7 \times \frac{1}{3} = 0.9 Degree C$

Answer 2

Answer:

Increase in the temperature of water would be 0.9 degree C

Explanation:

I got it right