# If a sample of chloroform is initially at 25°C, what is its final temperature of 150.0 g of chloroform absorbs 1.0 kilojoules of heat, and the specific heat of chloroform is 0.96 J/g°C?

##### 1 Answer

#### Answer:

#### Explanation:

**!! SHORT ANSWER !! **

This problem is a pretty straightforward application of the equation

#color(blue)(q = m * c * DeltaT)" "# , where

*final temperature* and the *initial temperature*

All you'd have to do here is rearrange this equation to solve for **do not** forget to convert the heat from *kilojoules* to *joules*

#DeltaT = q/(m * c)#

#DeltaT = (1.0 * 10^3 color(red)(cancel(color(black)("J"))))/(150.0 color(red)(cancel(color(black)("g"))) * 0.96color(red)(cancel(color(black)("J")))/(color(red)(cancel(color(black)("g"))) * ""^@"C")) = 6.94^@"C"#

The final temperature of the sample will thus be

#T_"final" = T_"initial" + DeltaT#

#T_"final" = 25^@"C" + 6.94^@"C"#

#T_"final" = color(green)(32^@"C") -># rounded to two sig figs

**!! AN ALTERNATIVE APPROACH !!**

I want to show you how to solve such questions **without** using that equation.

As you know, a substance's **specific heat** tells you how much heat is needed in order to increase the temperature of

In your case, chloroform is said to have a specific heat of

#c = 0.96"J"/("g" ""^@"C")#

*So, what does that tell you?*

In order to increase the temperature of

Now look at the data given to you. Focus on the *mass* of the sample first.

Now, if you need to provide

Since you need *per gram*, you will need

#150.0 xx "0.96 J" = "144 J"#

Notice that the problem tells you that you add **ratio** between the given heat and the amount of heat needed to increase the temperature of the sample by **increase in temperature**.

You can thus say that

#1.0 color(red)(cancel(color(black)("kJ"))) * (10^3 color(red)(cancel(color(black)("J"))))/(1 color(red)(cancel(color(black)("kJ")))) * (1^@"C")/(144 color(red)(cancel(color(black)("J")))) = 6.94^@"C"#

The final temperature of the sample will thus be

#T_"final" = 25^@"C" + 6.94^@"C" = 31.94^@"C"#

Rounded to two sig figs, the answer will once again be

#T_"final" = color(green)(32^@"C")#