91Ó°ÊÓ

A 0.50-g sample of vegetable oil is placed in a calorimeter with the following information: After the sample has been burnt,$18.9\mathrm{kJ}$is released. The oil's energy value (kcal/g) must be determined.

Calorimeter tests are used to determine the energy value of vegetable oil. The amount of energy in kilo joules or kilo calories is assessed in such tests by complete combustion of the fuel.$1\mathrm{g}$in the form of a vegetable oil The amount of energy obtained is equal to the energy value of vegetable oil.

Given:

Vegetable oil's mass is$0.50-\mathrm{g}$

Vegetable oil provides total energy of $18.9\mathrm{kJ}$

The energy value of the vegetable oil in question is a requirement.$\mathrm{kcal}/\mathrm{g}$

Plan:

When we look at the given value, total energy released for 0.50 - vegetable oil in gallons, we can see that$\mathrm{kJ}$, However, because the energy value of a grams of substance provides information, we must first determine the energy value in grams. $\mathrm{kJ}/\mathrm{g}$The energy value of vegetable oil can then be calculated using conversion factors.$\mathrm{kcal}/\mathrm{g}$.

Calculate the energy of the sample:

$18.9\mathrm{kJ}=0.50\mathrm{g}×$energy value in$\mathrm{kJ}$

energy value in $\mathrm{kJ}=\frac{18.9\mathrm{kJ}}{0.50\mathrm{g}}$

We must calculate the energy value in kcal.

Factors of equality and conversion:

$1\mathrm{kJ}=1000\mathrm{J}$

$1\mathrm{kcal}=1000\mathrm{cal}$

$1\mathrm{cal}=4.184\mathrm{J}$

The following conversion factors are related:

$\frac{1000\mathrm{J}}{1\mathrm{kJ}},\frac{1\mathrm{kcal}}{1000\mathrm{cal}},\frac{1\mathrm{cal}}{4.184\mathrm{J}}$respectively

We can now organise the energy value in $\frac{\mathrm{kJ}}{\mathrm{g}}$and related conversion factors in the following manner to obtain a unit response$\frac{kcal}{\mathrm{g}}$as follows:

$$\begin{gathered} \text{Energy value}=37.8\frac{\mathrm{N}}{\mathrm{g}}×\frac{1000)}{1\mathrm{k}}×\frac{1\mathrm{kcal}}{1000\mathrm{cal}}×\frac{1\mathrm{cal}}{4.184)} \\ =9.0344\frac{\mathrm{kcal}}{\mathrm{g}} \end{gathered}$$

Thus, energy value of the vegetable oll is $9.0344\mathrm{kcal}/\mathrm{g}$

Now we have to round off the result with proper significant figures, while solving problem we are using multiplications and divisions. In multiplication or in division the result will have same number of significant figures as the measurement with least significant figures.

If the conversion factor contains denominator and nominator units from same system then that division can be considered as exact value and we need not to consider significant figures.

If the conversion factor contains denominator and nominator units from same system then that division can be considered as exact value and we need not to consider significant figures.

The measurement$18.9\mathrm{J}$has three significant figures.

The measurement $0.50\mathrm{g}$has two significant figures.

The measurement $\frac{1000\mathrm{J}}{1\mathrm{kJ}}$is an exact value.

The measurement $\frac{1\mathrm{kcal}}{1000\mathrm{cal}}$is an exact value

The measurement $\frac{1\mathrm{cal}}{4.184\mathrm{J}}$is an exact value.

So the answer should contain two significant figures. Drop the final digits 344 to have the answer with two significant figures. The first digit to be dropped is 3 , which is less than 5 , hence we need not to increase one unit to the last retained number.

Thus, the energy value of vegetable oil with unit $\mathrm{kcal}/\mathrm{g}$$9.0\mathrm{kcal}/\mathrm{g}$