Problem 17.1

17.1
The container is thremally insulated, so there is no heat flowing out of the system. Q = 0
$\Delta U = Q - W$
$\Delta U = -W_{out} ~=~ W_{in}$
$\Delta U = 2mgh$
Let's find out how much this amount of work, in the form of heat energy would raise the temperature of the water.
$\Delta U = 2mgh = m_{water}c\Delta T$
$\Delta T = \frac{2mgh}{m_{water}c}$
$\Delta T = \frac{(2)(1.5)(9.8)(3)}{(.200)(4186)} ~=~ .105{}^{\circ}C$