# Question e1989

Apr 2, 2016

$1.67 \cdot {10}^{18} \text{J}$

#### Explanation:

The key to this problem is the specific heat of water, which is listed as

$\textcolor{p u r p \le}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{{c}_{\text{water" = "4.18 J g"^(-1)""^@"C}}^{- 1}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

Now, a substance's specific heat tells you how much heat is required in order to increase the temperature of $\text{1 g}$ of that substance by ${1}^{\circ} \text{C}$.

Water's specific heat tells you that you need $\text{4.18 J}$ of heat in order to increase the temperature of $\text{1 g}$ of water by ${1}^{\circ} \text{C}$.

In your case, you must increase the temperature $4.00 \cdot {10}^{11} {\text{m}}^{3}$ of water by

$\Delta T = {12.0}^{\circ} \text{C" - 11.0^@"C" = 1.0^@"C}$

This means that the the amount of heat needed will depend exclusively on the mass of water present in the lake.

If you assume that water has a density of ${\text{1000 kg m}}^{- 3}$, you can say that the volume of water present in the lake will amount to

4.00 * 10^(11)color(red)(cancel(color(black)("m"^3))) * "1000 kg"/(1color(red)(cancel(color(black)("m"^3)))) = 4.00 * 10^(14)"kg"

This will be equivalent to

4.00 * 10^(11)color(red)(cancel(color(black)("kg"))) * (10^3"g")/(1color(red)(cancel(color(black)("kg")))) = 4.00 * 10^(17)"g"#

So, to increase the temperature of water by ${1.0}^{\circ} \text{C}$, you need to supply $\text{4.18 J}$ for every gram you have in your sample, which means that you'll need

$4.00 \cdot {10}^{14} \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{g"))) * "4.18 J"/(1color(red)(cancel(color(black)("g")))) = color(green)(|bar(ul(color(white)(a/a)1.67 * 10^(18)"J} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

The answer is rounded to three sig figs.