Problem 17.8

17.8 a)
The mass of the water is
$m = (4 \times 10^{11} m^{3})(1000 kg/m^{3})$
So, the heat energy required to raise that amount of water $1{}^{\circ}C$ is
$\Delta Q = mc\Delta T$
$\Delta Q = (4 \times 10^{14} kg)(4186 J/m^{3})(1{}^{\circ}C)$
$\Delta Q = 1.68 \times 10^{18} J$

b)
The energy found in part a) can be expressed as power times time.
Q = Pt
Solving for the time,
$t = \frac{Q}{P}$
$t = \frac{1.68 \times 10^{18} J}{10^{9} J/s}$
or $t = 1.68 \times 10^{9} sec ~=~ 53.1$ years.