Question #3c48f

1 Answer
Dec 30, 2016

Answer:

Here's what I got.

Explanation:

The key here is the specific heat of water, which is usually given as

#c = "4.18 J g"^(-1)""^@"C"^(-1)#

The specific heat of a substance is the amount of heat that must be provided in order to increase the temperature of #"1 g"# of that substance by #1^@"C"#.

In your case, you need to provide #"4.18 J"# of heat in order to heat #"1 g"# of water by #1^@"C"#.

Now, the problem provides you the amount of heat that you have at your disposal and the desired change in temperature, which is equal to

#DeltaT = 75.0^@"C" - 20.0^@"C" = 55.0^@"C"#

Use the specific heat of water to calculate how much heat would be needed in order to increase the temperature of #"1 g"# of water by #55^@"C"#

#55.0 color(red)(cancel(color(black)(""^@"C"))) * "4.18 J"/("1 g" * 1color(red)(cancel(color(black)(""^@"C")))) = "229.9 J g"^(-1)#

So, you must add #"229.9 J"# for every #"1 g"# of water in order to cause its temperature to increase by #55.0^@"C"#. This means that #"8000 J"# of heat will get you

#8000 color(red)(cancel(color(black)("J"))) * overbrace("1 g"/(229.9 color(red)(cancel(color(black)("J")))))^(color(blue)("needed for 55" ""^@"C increase")) = color(darkgreen)(ul(color(black)("35.8 g")))#

I'll leave the answer rounded to three sig figs, but keep in mind that you only provided one significant figure for the amount of heat you have at your disposal.