# The specific heat of liquid bromine is 0.226J/g-K and the density is 3.12g/mL, how much heat (J) is required to raise the temperature of 10.0mL of bromine from 25.00 C to 27.30 C?

Nov 14, 2015

$\text{16.2 J}$

#### Explanation:

The idea here is that you need to use bromine's density to determine how many grams of bromine you have in that $\text{10.0-mL}$ sample.

A density of $\text{3.12 g/mL}$ tells you that you get a mass of $\text{3.12 g}$ for every $\text{100 mL}$ of bromine. This means that your sample will have a mass of

10.0color(red)(cancel(color(black)("mL"))) * "3.12 g"/(1color(red)(cancel(color(black)("mL")))) = "31.2 g"

Now, the equation that establishes a relationship between heat absorbed and change in temperature looks like this

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

$q$ - heat absorbed
$m$ - the mass of the sample
$c$ - the specific heat of the substance
$\Delta T$ - the change in temperature, defined as final temperature minus initial temperature

Plug in your values and solve for $q$ to get

$q = 31.2 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{g"))) * 0.226"J"/(color(red)(cancel(color(black)("g"))) color(red)(cancel(color(black)(""^@"C")))) * (27.30 - 25.00)color(red)(cancel(color(black)(""^@"C}}}}$

$q = \textcolor{g r e e n}{\text{16.2 J}}$