# Question ce46f

Feb 1, 2017

${33.3}^{\circ} \text{C}$

#### Explanation:

The key to this problem is the specific heat of olive oil, which is said to be equal to

${c}_{\text{olive oil" = "2.19 cal g"^(-1)""^@"C}}^{- 1}$

As you know, the specific heat of a substance tells you how much heat must be added to $\text{1 g}$ of that substance in order to increase its temperature by ${1}^{\circ} \text{C}$.

In your case, you must add $\text{2.19 cal}$ of heat for every $\text{1 g}$ of olive oil in order to increase its temperature by ${1}^{\circ} \text{C}$.

Now, the first thing to do here is to figure out how much heat is needed in order to increase the temperature of $\text{55.0 g}$ of oil by ${1}^{\circ} \text{C}$

55.0 color(red)(cancel(color(black)("g"))) * "2.19 cal"/(1color(red)(cancel(color(black)("g"))) * 1^@"C") = "120.45 cal"^@"C"^(-1)

This means that in order to increase the temperature of $\text{55.0 g}$ of olive oil by ${1}^{\circ} \text{C}$, you need to supply it with $\text{120.45 cal}$ of heat.

Consequently, adding $\text{877 cal}$ of heat will produce an increase in temperature of

877 color(red)(cancel(color(black)("cal"))) * overbrace( (1^@"C")/(120.45color(red)(cancel(color(black)("cal")))))^(color(blue)("for 55.0g of olive oil")) = 7.28^@"C"#

Therefore, the final temperature of the olive oil will be

$\textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{{T}_{\text{final" = 26.0^@"C" + 7.28^@"C" = 33.3^@"C}}}}}$

The answer is rounded to three sig figs.