## College Physics (4th Edition)

(a) The internal energy increases by $34.3~J$ (b) It is not likely that the water temperature increases.
(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.