Question

To get an idea of how much thermal energy is contained in the world’s oceans, estimate the heat liberated when a cube of ocean water, 1 km on each side, is cooled by 1 K. (Approximate the ocean water as pure water for this estimate.)

Short answer

The liberated heat is $4.2\times 1015\mathrm{J}$.

Step-by-step solution

Given data

The water cube has the side length$L=1\mathrm{k}\mathrm{m}=103\mathrm{m}$.

The temperature decrease of the water is $\mathrm{\Delta }T=1\mathrm{K}=1\circ \mathrm{C}$.

Concept

The density of the water is$\rho =1000\mathrm{k}\mathrm{g}/\mathrm{m}3$.

In this problem, find the mass of the water first.

Calculation of part (a)

The specific heat of the water is $c=4186\mathrm{J}/\mathrm{k}\mathrm{g}\cdot \circ \mathrm{C}$.

The volume of the cube of water is $V=L3$.

The mass of the water is

$\begin{array}{c}m=V\rho \\ =L3\rho \\ =\left(103m\right)3\times \left(1000\mathrm{k}\mathrm{g}/\mathrm{m}3\right)\\ =1012\mathrm{k}\mathrm{g}.\end{array}$

Now, the heat liberated when the water is cooled by 1 K is

$\begin{array}{c}Q=mc\mathrm{\Delta }T\\ =\left(1012\mathrm{k}\mathrm{g}\right)\times \left(4186\mathrm{J}/\mathrm{k}\mathrm{g}\cdot \circ \mathrm{C}\right)\times \left(1\circ \mathrm{C}\right)\\ =4.2\times 1015\mathrm{J}.\end{array}$

Hence, the liberated heat is $4.2\times 1015\mathrm{J}$.