A paddle wheel is coupled to a mass of 12 kg via fixed frictionless pulleys. The paddle is immersed in a liquid of heat capacity 4200 J K ⁻¹ enclosed in an adiabatic container. Suppose that during a time interval, the 12 kg block falls slowly through 70 cm . Calculate the work done on the liquid.

Step-by-step solution

The work done on the liquid by the paddle wheel is equal to the decrease in the potential energy of the falling mass. Since the block falls slowly, there is no change in its kinetic energy. The work done is given by: W = mgh Given: Mass of the block, m = 12 kg Distance fallen, h = 70 cm = 0.7 m Acceleration due to gravity, g = 10 m/s ^2 Substituting the values: W = 12 10 0.7 W = 84 J