Question #71e95

1 Answer
Apr 3, 2016

Answer:

#58.2^@"C"#

Explanation:

In order to be able to solve this problem, you need to know the value of water's specific heat, which you'll find 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)|)))#

A substance's specific heat tells you how much heat is needed in order to increase the temperature of #"1 g"# of that substance by #1^@"C"#.

In your case, you can say that in order to increase the temperature of #"1 g"# of water by #1^@"C"#, you need to provide it with #"4.18 J"# of heat.

Keep in mind that this much heat must be applied per gram, per degree Celsius.

Now, you know that your sample of water has a mass of #"150. g"#, and that it absorbs a total of #"21.8 kJ"# of heat. Convert the heat from kilojoules to joules by using the conversion factor

#"1 kJ" = 10^3"J"#

to get

#21.8color(red)(cancel(color(black)("kJ"))) * (10^3"J")/(1color(red)(cancel(color(black)("kJ")))) = "21,800 J"#

At this point, you can determine how much heat would be needed in order to increase the temperature of this sample of water by #1^@"C"#. Well, you know that you need #"4.18 J"# per gram in order to get a #1^@"C"# increase in temperature, so #"150. g"# of water will require

#150. color(red)(cancel(color(black)("g"))) * overbrace("4.18 J"/(1color(red)(cancel(color(black)("g")))))^(color(purple)("heat needed for 1"^@"C increase for 1 g")) = "627 J"#

If you need #"627 J"# of heat in order to increase the temperature of #"150 g"# of water by #1^@"C"#, it follows that the difference between this much heat and the #"21,800 J"# you supplied went in increasing the sample's temperature.

You can thus say that

#"21,800"color(red)(cancel(color(black)("J"))) * overbrace((1^@"C")/(627color(red)(cancel(color(black)("J")))))^(color(blue)("heat needed for 1"^@"C for 150. g")) = 34.77^@"C"#

If the sample started at an initial temperature of #23.4^@"C"#, you can say that its final temperature will be

#T_"final" = 23.4^@"C" + 34.77^@"C" = color(green)(|bar(ul(color(white)(a/a)58.2^@"C"color(white)(a/a)|)))#

The answer is rounded to three sig figs.

ALTERNATIVE APPROACH

You can get the same result by suing the equation

#color(blue)(|bar(ul(color(white)(a/a)q = m * c * DeltaTcolor(white)(a/a)|)))" "#, where

#q# - the amount of heat gained / lost
#m# - the mass of the sample
#c# - the specific heat of the substance
#DeltaT# - the change in temperature, defined as the difference between the final temperature and the initial temperature

In your case, you'd have

#DeltaT = q/(m * c)#

#DeltaT = ("21,800" color(red)(cancel(color(black)("J"))))/(150. color(red)(cancel(color(black)("g"))) * 4.18color(red)(cancel(color(black)("J")))color(red)(cancel(color(black)("g"^(-1))))""^@"C"^(-1)) = 34.77^@"C"#

Once again, you'd have

#T_"final" = 23.4^@"C" + DeltaT = color(green)(|bar(ul(color(white)(a/a)58.2^@"C"color(white)(a/a)|)))#