Question

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?

Answer 1

`32^@"C"`

Explanation:

!! SHORT ANSWER !!

This problem is a pretty straightforward application of the equation

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

`q` - the amount of heat added / removed
`m` - the mass of the sample
`c` - the specific heat of the substance
`DeltaT` - the change in temperature, calculated as the difference between the final temperature and the initial temperature

All you'd have to do here is rearrange this equation to solve for `DeltaT` and plug in your values - 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 `"1 g"` of that substance by `1^@"C"`.

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 `"1 g"` of chloroform by `1^@"C"`, you need to provide it with `"0.96 J"` of heat.

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

Now, if you need to provide `"0.96 J"` of heat in order to increase the temperature of `"1 g"` of chloroform by `1^@"C"`, how much heat will be needed in order to increase the temperature of `"150.0 g"` of chloroform by `1^@"C"`?

Since you need `"0.96 J"` per gram, you will need

`150.0 xx "0.96 J" = "144 J"`

Notice that the problem tells you that you add `"1 kJ"` of heat to the sample. This means that the ratio between the given heat and the amount of heat needed to increase the temperature of the sample by `1^@"C"` will represent the actual 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")`