# 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

#### Explanation:

A substance's **specific heat** tells you the amount of heat needed to increase the temperature of

In your case, you know that the temperature of

The first thing to do here is to figure out how much heat is needed to increase the temperature of

#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

#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**.

**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

*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