A 0.050g ball of aluminum foil absorbs 1.42 J heat when it is heated 31.6°C. What is the specific heat of aluminum foil?

1 Answer
Aug 13, 2016

Answer:

#"0.899 J g"^(-1)""^@"C"^(-1)#

Explanation:

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

In your case, you know that the temperature of #"0.050 g"# of aluminium increased by #31.6^@"C"# upon the absorption of #"1.42 J"# of heat.

The first thing to do here is to figure out how much heat is needed to increase the temperature of #"1 g"# of aluminium by #31.6^@"C"#.

#1 color(red)(cancel(color(black)("g"))) * "1.41 J"/(0.050color(red)(cancel(color(black)("g")))) = "28.4 J"#

This is how much heat is needed to cause a #31.6^@"C"# increase in the temperature of #"1 g"# of aluminium. Now all you have to do is calculate how much heat is needed to increase the temperature of #"1 g"# by #1^@"C"#

#1 color(red)(cancel(color(black)(""^@"C"))) * "28.4 J"/(31.6color(red)(cancel(color(black)(""^@"C")))) = "0.899 J"#

Therefore, the specific heat of aluminium will be

#color(green)(|bar(ul(color(white)(a/a)color(black)(c_"Al" = "0.899 J g"^(-1)""^@"C"^(-1))color(white)(a/a)|)))#

The answer is rounded to three sig figs.

#color(white)(a/a)#

ALTERNATIVELY

You can get the same result by suing the following equation

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

#q# - the amount of heat released
#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, the change in temperature is equal to

#DeltaT = 31.6^@"C"#

You will have

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

Plug in your values to find

#c_"Al" = "1.42 J"/("0.050 g" * 31.6^@"C") = color(green)(|bar(ul(color(white)(a/a)color(black)("0.899 J g"^(-1)""^@"C"^(-1))color(white)(a/a)|)))#

The listed value for the specific heat of aluminium is

#c_"Al" = "0.900 J g"^(-1)""^@"C"^(-1)#

http://hyperphysics.phy-astr.gsu.edu/hbase/tables/sphtt.html