Question

Figure (26-E1) shows a paddle wheel coupled to a mass of 12kg through fixed frictionless pulleys. The paddle is immersed in a liquid of heat capacity 4200J/K kept in an adiabatic container. Consider a time interval in which the 12kg 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 1

(a) Heat is not given to the liquid. Instead the mechanical work done is converted to heat. So, heat given to liquid is z.
(b) Work done on the liquid is the PE lost by the 12kg mass = mgh = 12 × 10× 0.70 = 84J

(c) Rise in temp at Δt 

We know, 84 = msΔt
=> 84 = 1 × 4200 × Δt (for ‘m’ = 1kg) 

=> Δt = 84/4200 = 0.02k