10 l of water is dropped from a height of 250 m. How much heat (In calories) will be generated when the water reaches the bottom? Assuming that the entire heat will be retained by the mass of water, what will be the rise in temperature of the water? (Given `J = 4.18 J cdot cal^(-1))`

Explanation

Mass of 10 l of water, m = 10 kg
Kinetic energy on impact with the ground
= initial potential energy = mgh = work done (W)
So, heat generated,
`H = W/J = (mgh)/(J) cal`
The specific heat of water,
`s = 1000 cal cdot kg^(-1) .^(@)C^(-1)`,
If t is the rise in temperature, then
`mst = H = (mgh)/(J)`
or, `t = (gh)/(Js) = (9.8 xx 250)/(4.18 xx 1000) = 0.586^(@)C`.