# 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?

Jan 23, 2016

${32}^{\circ} \text{C}$

#### Explanation:

This problem is a pretty straightforward application of the equation

$\textcolor{b l u e}{q = m \cdot c \cdot \Delta T} \text{ }$, where

$q$ - the amount of heat added / removed
$m$ - the mass of the sample
$c$ - the specific heat of the substance
$\Delta T$ - 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 $\Delta T$ and plug in your values - do not forget to convert the heat from kilojoules to joules

$\Delta T = \frac{q}{m \cdot c}$

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

${T}_{\text{final" = T_"initial}} + \Delta T$

${T}_{\text{final" = 25^@"C" + 6.94^@"C}}$

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 $\text{1 g}$ of that substance by ${1}^{\circ} \text{C}$.

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 $\text{1 g}$ of chloroform by ${1}^{\circ} \text{C}$, you need to provide it with $\text{0.96 J}$ of heat.

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

Now, if you need to provide $\text{0.96 J}$ of heat in order to increase the temperature of $\text{1 g}$ of chloroform by ${1}^{\circ} \text{C}$, how much heat will be needed in order to increase the temperature of $\text{150.0 g}$ of chloroform by ${1}^{\circ} \text{C}$?

Since you need $\text{0.96 J}$ per gram, you will need

$150.0 \times \text{0.96 J" = "144 J}$

Notice that the problem tells you that you add $\text{1 kJ}$ 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 ${1}^{\circ} \text{C}$ 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

${T}_{\text{final" = 25^@"C" + 6.94^@"C" = 31.94^@"C}}$

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

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