Question

The specific heat of the liquid bromine is $0.226{\text{ }}\dfrac{j}{{gK}}$ and the density is$3.12{\text{ }}\dfrac{g}{{mL}}$. How much heat (J) is required to raise the temperature of $10.0{\text{ }}mL$of bromine from$25.00{\text{ }}C{\text{ }}to{\text{ }}27.30{\text{ }}C$?

Answer 1

The specific heat capacity of a substance is the heat required to raise the temperature of the unit mass of a given substance by a given amount. Use the formula of specific heat capacity to calculate the amount of heat. Formula: $q = mc\Delta T$
($q$ - heat absorbed or amount of heat,$m$ - the mass of the sample,$c$ - the specific heat capacity of the substance and$\Delta T$ - the change in temperature $\left( {{T_2} - {T_1}} \right)$

Complete answer:
In the above question, you are given the specific heat capacity and density of bromine. We have to find the amount of heat required to raise the temperature of $10.0{\text{ }}mL$of bromine. For finding the heat the formula of specific heat capacity will be used. The formula is:
$q = mc\Delta T$
where$q$ - heat absorbed or amount of heat,$m$ - the mass of the sample,$c$ - the specific heat capacity of the substance and $\Delta T$ - the change in temperature $\left( {{T_2} - {T_1}} \right)$
For this, we need to find out the mass of the sample and in the question, the mass of the sample is not given so we use the density of the bromine to find out the mass of the sample.
It is given that the density of liquid bromine is $3.12{\text{ }}g{m^{ - 1}}{L^{ - 1}}$. This means that a mass of $3.12{\text{ }}g$ of sample for every $100{\text{ }}mL\;$ of liquid bromine. So the mass of the sample will turn out to be:
Thus the mass of the sample is $31.2g$.
Now use this mass to calculate the amount of heat. The value of $q$ you have to calculate, $m = 31.2g$, $c = 0.226\dfrac{j}{{gK}}$, $\Delta T = ({T_1} = {25.0^0}C,{\text{ }}{T_2} = {27.30^0}C)$. Substitute these values in the formula and calculate.
$q = mc\Delta T$
$\Rightarrow q = m \times c \times \left( {{T_2} - {T_1}} \right)$
$\Rightarrow q = 31.2 \times 0.226 \times \left( {27.30 - 25.0} \right)$
$\Rightarrow q = 31.2 \times 0.226 \times 2.3$
$\Rightarrow q = 16.21776{\text{ }}J$
Therefore the amount of heat required to raise the temperature is $16.21776{\text{ }}J$

Note:
The simple idea to solve this problem is to use bromine's density to determine how many grams of bromine you have in that $10.0{\text{ }}mL$ sample. The temperature $\Delta T$ is the change in temperature that is the final temperature minus the initial temperature. Solving this problem directly might seem difficult as the mass is not given, that is why to observe the questions and think before solving.