Question #de7a7

1 Answer
Sep 17, 2017

Answer:

#140^@"C"#

Explanation:

The key here is the specific heat of glass because it tells you the amount of heat needed in order to increase the mass of #"1 g"# of glass by #1^@"C"#.

#c_"glass" = "0.50 J g"^(-1)""^@"C"^(-1)#

So, you know that in order to increase the temperature of #"1 g"# of glass by #1^@"C"#, you need to supply it with #"0.50 J"# of energy.

Now, your sample has a mass of #"103.2 g"#. Use the specific heat of glass to calculate how much heat is needed to increase the temperature of this sample

#103.2 color(red)(cancel(color(black)("g"))) * overbrace("0.50 J"/(1 color(red)(cancel(color(black)("g"))) * 1^@"C"))^(color(blue)("the specific heat of glass")) = "51.6 J"""^@"C"^(-1)#

This tells you that in order to increase the temperature of #"103.2 g"# of glass by #1^@"C"#, you need to provide #"51.6 J"#.

In your case, #"6375 J"# will increase the temperature of #"103.2 g"# of glass by

#6375 color(red)(cancel(color(black)("J"))) * (1^@"C")/(51.6 color(red)(cancel(color(black)("J")))) = 123.55^@"C"#

At this point, you should round the value to two sig figs because that's how many sig figs you have for the specific heat of glass.

So you will have

#123.55^@"C" ~~ 120^@"C"#

Since this represents the increase in temperature that corresponds to a #"103.2-g"# sample of glass that absorbs #"6375 J"#, you can say that the final temperature of the glass will be

#T_"final" = 20.0^@"C" + 120^@"C" = color(darkgreen)(ul(color(black)(140^@"C")))#

The answer cannot have any decimal places because #120^@"C"# does not have any decimal places.