Question #5dc43

1 Answer
Mar 20, 2016

Answer:

#"0.015 kg Al"#

Explanation:

The key to this problem is aluminium's specific heat, which tells you how much heat is needed to increase the temperature of #"1 g"# of aluminium by #1^@"C"#.

Aluminium's specific heat is said to be equal to #"0.897 J g"^(-1)""^@"C"^(-1)#. This tells you that in order to increase the temperature of #"1 g"# of aluminium by #1^@"C"#, you need to supply it with #"0.897 J"# of heat.

Now, let's assume that your sample has a mass of #"1 g"#. In order to increase its temperature by

#DeltaT = 187^@"C" - 65^@"C" = 122^@"C"#

you would need to provide it with #122# times more heat than what would be needed to increase its temperature by #1^@"C"#.

More specifically, you would need to provide it with

#122color(red)(cancel(color(black)(""^@"C"))) * overbrace("0.897 J"/(1color(red)(cancel(color(black)(""^@"C")))))^(color(purple)("heat needed for"color(white)(a)1^@"C"color(white)(a) "increase for 1 g of Al")) = "109.434 J"#

However, you end up adding #"1650 J"# of heat to this sample, which of course means that its mass is not equal to #"1 g"#.

The ratio between the heat added to the sample to get its temperature to increase by #122^@"C"# and the heat that would be needed to increase the temperature of #"1 g"# of aluminium by #122^@"C"# will give you the mass of the sample, expressed in grams

#1650color(red)(cancel(color(black)("J"))) * overbrace("1 g"/(109.434color(red)(cancel(color(black)("J")))))^(color(purple)("heat needed for"color(white)(a) 122^@"C" color(white)(a)"increase for 1 g of Al")) = "15.08 g"#

Expressed in kilograms and rounded to two sig figs, the answer will be

#color(purple)(|bar(ul(color(white)(a/a)color(black)("1 kg" 1= 10^3"g")color(white)(a/a)|)))#

#"mass of Al" = color(green)(|bar(ul(color(white)(a/a)"0.015 kg"color(white)(a/a)|)))#

ALTERNATIVE SOLUTION

You can get the same result by using the formula

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

#q# - the amount of heat added
#m# - the mass of the sample
#c# - the specific heat of the substance
#DeltaT# - the change in temperature

Rearrange to solve for #m#

#q = m * c * DeltaT implies m = q/(c * DeltaT)#

Plug in your values to get

#m = (1650 color(red)(cancel(color(black)("J"))))/(0.897color(red)(cancel(color(black)("J"))) "g" color(red)(cancel(color(black)(""^@"C"^(-1)))) * (187 - 65)color(red)(cancel(color(black)(""^@"C"))))#

#m = "15.08 g"#

Once again, you'll have

#"mass of Al" = color(green)(|bar(ul(color(white)(a/a)"0.015 kg"color(white)(a/a)|)))#