Chapter 14, Problem 47GP

To determine

To find : Heat liberated by cube of ocean water.

Answer to Problem 47GP

  $4.2\times {10}^{15}$ Joule

Explanation of Solution

Given :

Side of melted cube $=1$ km

  $\Delta T=1\text{K}$

  $S=1\text{ kcal/kg }\cdot {\text{k}}^{-1}$

Formula Used :

Heat is given as,

  $Q=ms\Delta T$

Calculation :

Since $m=\rho v$

So,

  $\begin{array}{c}Q=ms\Delta T\\ =\rho vs\Delta T\end{array}$

Substituting the vales,

  $\Rightarrow Q=1000\times {\left(1000\right)}^3\times 1\times 1$

  $\Rightarrow Q={10}^{12}$ kcal

  $\Rightarrow Q=4.2\times {10}^{15}$ Joule

Conclusion :

Hence $4.2\times {10}^{15}$ Joule of heat will be liberated.