Shows a peddle wheel couple to a mass of 12 kg through fixed frictionless pulley. The paddle is immersed in a liquid of heat capacity `4200 J K^(-1)` kept in an adiabatic container. Consider a time interval in which the 12 kg block falls slowly through 70cm. (a) how much heat is given to the liquid ? (b) how much work is done on the liquid? (c ) calculate the rise in the temperature of the liquid neglecting the heat capacity of the container and the paddle.

Answer

A, B, C, D

Explanation

(a) heat is not given to the liqiud Instead, the mechanical workdone is converted to heat.
So, heat given to liquid is zero
(b) Work done on the liquid is the PE lost by the 12 kg mass =mgh`
`=12x10x0*70ltbrge84 J
(c ) Let rise in temperature be Deltat We know, 84=`msDeltat,`
`rArr 84=1xx4200xxDeltat(for m =1 kg)`
` rArr Deltat=(84)/(4200)=(1)/(50)=0*02 k.