On a winter day temperature of the tap water is `20^(@)C` where as the room temperature is `5^(@)C` water is stored in a tank of capacity `0.5 m^(3)` for household use .If it were possible to use the beat liberated by the water to lift a `10kg` mass vertically , how high can it be lifted as the water comes to the room temperature ? Take `g = 10ms^(-2)`

Answer

A, C

Explanation

Here `m == 0.5m^(2) = 500L = 500kg`
So the heat liberated during the water changes`20^(@)C to 5^(@)C`
`= 500 xx 4200 xx 15`[Delta theta = 20- 5= 15]`
`= 500 xx 4200 xx 15`
`= 75 xx 420 xx 1000`
`= 31500 xx 1000`
Let the height = h`
The required work
`= mgh = 10 xx 10 xx h= 100h`
But,`100h = 315000m`
`rArr h = 31500 m = 315km`