Question

Figure shows a paddle wheel coupled to a mass of 12 kg through fixed frictionless pulleys. The paddle is immersed in a liquid of heat capacity $4200\text{J}{K}^{-1}$ kept in an adiabatic container. Consider a time interval in which the 12 kg block falls slowly through 70 cm.

(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 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

$=12\times 10\times 0.70$

$=84\text{J}$

(c) Let rise in temperature be $\Delta t$

We know, $84=ms\Delta t$

$\Rightarrow 84=1\times 4200\times \Delta t$ (for 'm' = 1 kg)

$\Delta t=\frac{84}{4200}=\frac{1}{50}=0.02\text{K}$