One kg of liquid water at 25°C, for which

Question

(a) Experiences a temperature increase of 1 K. What is , in kJ?

(b) Experiences a change in elevation ?z. The change in potential energy is the same as for part (a). What is ?z, in meters?

(c) Is accelerated from rest to final velocity u. The change in kinetic energy is the same as for part (a). What is u, in

Answer 1

In the previous problems, the heat capacity of water was given as For liquids, and are very

nearly the same, so it doesn’t matter whether this change occurs at constant P or at constant V. The change in internal energy is thus



a. The change in gravitational potential energy is

= mg?z

So, to have the same change in potential energy as the change in internal energy in part (a), we need

4180 J =

from which ?z = 427 m (or 1400 feet)

b. This time, we need

= 1?2mv 2 = 4180 J, from which or v = 91.4 m/s (or 205 miles per hour).

What have we learned? We have discovered again that what corresponds, in our everyday experience, to a small amount of thermal energy corresponds to a relatively large amount of mechanical energy. The amount of energy required to heat a quart of water (1 kg of water is about 1 liter or 1 quart) from room temperature to its boiling point (increasing the temperature by about 75 K) would be enough to raise the water to a height of about 20 miles (in a frictionless world, or one in which we lifted the object very slowly so that the drag of the air on it was negligible).