# Question #2c289

Mar 3, 2017

$\text{23,000 J}$

#### Explanation:

Your tool of choice here will be the equation

$\textcolor{b l u e}{\underline{\textcolor{b l a c k}{q = m \cdot c \cdot \Delta T}}}$

Here

• $q$ is the heat lost or gained by the substance
• $m$ is the mass of the sample
• $c$ is the specific heat of the substance
• $\Delta T$ is the change in temperature, defined as the difference between the final temperature and the initial temperature of the sample

In your case, the change in temperature is equal to ${66}^{\circ} \text{C}$. In other words, the difference between the final and the initial temperatures of the water is equal to ${66}^{\circ} \text{C}$.

Now, look up the specific heat for water. You should find it listed as

${c}_{\text{water" = "4.18 J g"^(-1)""^@"C}}^{- 1}$

The specific heat of a substance is a measure of how much energy is needed to increase the temperature of $\text{1 g}$ of said substance by ${1}^{\circ} \text{C}$.

So, plug in your values to find

$q = 83 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{g"))) * "4.18 J" color(red)(cancel(color(black)("g"^(-1)))) color(red)(cancel(color(black)(""^@"C"^(-1)))) * 66color(red)(cancel(color(black)(""^@"C}}}}$

$q = \textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{\text{23,000 J}}}}$

The answer must be rounded to two sig figs, the number of significant figures you have for the mass of the sample and for the change in temperature.