Answer
(a) The internal energy increases by $34.3~J$
(b) It is not likely that the water temperature increases.
Work Step by Step
(a) The increase in internal energy is equal to the initial gravitational potential energy of the water at a height of 2.5 meters:
$\Delta E = mgh$
$\Delta E = (1.4~kg)(9.80~m/s^2)(2.5~m)$
$\Delta E = 34.3~J$
The internal energy increases by $34.3~J$
(b) We can calculate the change in temperature of the water:
$\Delta T = \frac{Q}{m~c}$
$\Delta T = \frac{34.3~J}{(6.4~kg)(4186~J/kg~K)}$
$\Delta T = 0.0013~K$
Since the expected change in temperature is so small, it is not likely that the water temperature increases.
