Through fixed frictionless pulleys, a paddle wheel is connected to a 12 kg mass. The paddle is immersed in a liquid of heat capacity 4200 J K ⁻¹ located in an adiabatic container. Consider a time interval in which the 12 kg block drops slowly by 70 cm . Determine the rise in the temperature of the liquid, assuming the heat capacity of the container and the paddle is negligible.

Step-by-step solution

The loss in potential energy of the block is converted into the internal energy of the liquid. Since the block drops slowly, the change in its kinetic energy is negligible. Work done by gravity on the block is: W = mgh Substituting the given values ( m = 12 kg , g = 10 m s ⁻² , h = 0.7 m ): W = 12 10 0.7 = 84 J This work done increases the temperature of the liquid. Let C be the heat capacity of the liquid and T be the rise in temperature: W = C T Given C = 4200 J K ⁻¹ : 84 = 4200 T T = 84 4200 = 0.02 K A temperatu