# Question 779cf

Mar 1, 2017

$\text{19,500 J}$

#### Explanation:

In order to be able to solve this problem, you must know the specific heat of water, which you'll find listed as

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

Now, the specific heat of a given substance tells you the amount of heat needed to increase the temperature of $\text{1 g}$ of said substance by ${1}^{\circ} \text{C}$.

In the case of water, you have

c_"water" = color(blue)("4.18 J") color(white)(.)color(red)("g"^(-1) color(white)(.)color(purple)(""^@"C"^(-1))

which means that you need $\textcolor{b l u e}{\text{4.18 J}}$ of heat to raise the temperature of $\textcolor{red}{\text{1 g}}$ of water by ${1}^{\circ} \text{C}$.

Now, you need to raise the temperature of $\text{85.0 g}$ of water by

$\Delta T = {80.0}^{\circ} \text{C" - 25.0^@"C" = 55.0^@"C}$

The first thing to do here is to calculate how much heat will be needed to raise the temperature of $\text{85.0 g}$ of water by ${1}^{\circ} \text{C}$. To do that, use the specific heat of water as a conversion factor

85.0 color(red)(cancel(color(black)("g"))) * "4.18 J"/(1color(red)(cancel(color(black)("g"))) * 1^@"C") = "355.3 J" ""^@"C"^(-1)#

This means that for every ${1}^{\circ} \text{C}$ increase in the temperature of an $\text{85.0-g}$ sample of water, you need $\text{355.3 J}$ of heat.

In your case, you will need

$55.0 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{^@"C"))) * "355.3 J"/(1color(red)(cancel(color(black)(""^@"C")))) = color(darkgreen)(ul(color(black)("19,500 J}}}}$

The answer is rounded to three sig figs.

You can double-check this result by using 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, you will have

$q = 85.0 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{g"))) * "4.18 J" color(red)(cancel(color(black)("g"^(-1)))) color(red)(cancel(color(black)(""^@"C"^(-1)))) * 55.0 color(red)(cancel(color(black)(""^@"C}}}}$

$q = \textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{\text{19,500 J}}}}$