# Question #6ad9d

##### 1 Answer

#### Answer:

#### 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

In your case, you know that you need

You can use the specific heat of aluminium to calculate how much heat would be required to raise the temperature of

#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

#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**.

**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 thechange in temperature, defined as the difference between thefinal temperatureand theinitial 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")))#