Through fixed frictionless pulleys, a paddle wheel is connected to...

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 ($m = 12 \text{ kg}$, $g = 10 \text{ m s}^{-2}$, $h = 0.7 \text{ m}$):

$W = 12 \times 10 \times 0.7 = 84 \text{ J}$

This work done increases the temperature of the liquid. Let $C$ be the heat capacity of the liquid and $\Delta T$ be the rise in temperature:

$W = C \Delta T$

Given $C = 4200 \text{ J K}^{-1}$:

$84 = 4200 \times \Delta T$

$\Delta T = \dfrac{84}{4200} = 0.02 \text{ K}$

A temperature change of $0.02 \text{ K}$ is equivalent to a change of 0.02^{}^{{}^{\circ}} \text{C}.