# Question #e1989

##### 1 Answer

#### Explanation:

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

#color(purple)(|bar(ul(color(white)(a/a)color(black)(c_"water" = "4.18 J g"^(-1)""^@"C"^(-1))color(white)(a/a)|)))#

Now, a substance's specific heat tells you how much heat is required in order to increase the temperature of

Water's specific heat tells you that you need

In your case, you must increase the temperature

#DeltaT = 12.0^@"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

#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 **for every gram** you have in your sample, which means that you'll need

#4.00 * 10^(14)color(red)(cancel(color(black)("g"))) * "4.18 J"/(1color(red)(cancel(color(black)("g")))) = color(green)(|bar(ul(color(white)(a/a)1.67 * 10^(18)"J"color(white)(a/a)|)))#

The answer is rounded to three **sig figs**.