Question #6ad9d

1 Answer
Dec 11, 2016

Answer:

#"2250 J"#

Explanation:

The key to this problem is the specific heat of aluminium

#c_"Al" = color(purple)("0.899 J")/(color(green)("g")color(blue)(""^@"C"))#

The specific heat of a substance tells you how much heat is required to reaise the temperature of #"1 g"# of said substance by #1^@"C"#.

In your case, you know that you need #color(purple)("0.899 J"# of heat in order to increasde the temperature of #color(green)("1 g"# of aluminium by #color(blue)(1^@"C")#.

You can use the specific heat of aluminium to calculate how much heat would be required to raise the temperature of #"500.0 g"# of aluminium by #1^@"C"#

#500.0 color(red)(cancel(color(black)("g"))) * color(purple)("0.899 J")/(color(green)(1)color(red)(cancel(color(green)("g"))) color(blue)(1^@"C")) = "449.5 J"color(blue)(""^@"C"^(-1))#

This tells you that you need to provide #"449.5 J"# in order to raise the heat of #"500.0 g"# of aluminium by #1^@"C"#. The problem tells you that the temperature of the sample must increase by

#DeltaT = 20.0^@"C" - 15.0^@"C" = 5.0^@"C"#

In order to get thatto happen, you must provide it with

#5.0 color(red)(cancel(color(black)(""^@"C"))) * "449.5 J"/(color(blue)(1)color(red)(cancel(color(blue)(""^@"C")))) = ul("2250 J")#

The answer is rounded to three sig figs.

#color(white)(a)#
ALTERNATIVELY

you can also use the following equation

#color(blue)(ul(color(black)(q = m * c * DeltaT)))#

Here

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

In your case, you will once again end up with

#q = 500.0 color(red)(cancel(color(black)("g"))) * "0.899 J" color(red)(cancel(color(black)("g"^(-1)))) color(red)(cancel(color(black)(""^@"C"^(-1)))) * (20.0 - 15.0) color(red)(cancel(color(black)(""^@"C")))#

#q = color(darkgreen)(ul(color(black)("2250 J")))#