Question #e1989

1 Answer
Apr 2, 2016

Answer:

#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.