Question

Question #e1989

Answer 1

`1.67 * 10^(18)"J"`

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 `"1 g"` of that substance by `1^@"C"`.

Water's specific heat tells you that you need `"4.18 J"` of heat in order to increase the temperature of `"1 g"` of water by `1^@"C"`.

In your case, you must increase the temperature `4.00 * 10^(11)"m"^3` of water by

`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 `"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^@"C"`, you need to supply `"4.18 J"` 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.