Question 4113d

Aug 21, 2017

$\text{9,500 J}$

Explanation:

An important assumption to make here is that you're increasing the temperature of liquid water by ${87}^{\circ} \text{C}$. In other words, you should assume that this increase in temperature does not include a phase change.

Now, the key here is water's specific heat

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

This tells you the amount of energy needed to increase the temperature of $\text{1 g}$ of water by ${1}^{\circ} \text{C}$.

So, for a sample of $\text{1-g}$ of water, you need $\text{4.18 J}$ of heat to increase its temperature by ${1}^{\circ} \text{C}$. This means that for any sample of water, you will need

87 color(red)(cancel(color(black)(""^@"C"))) * overbrace("4.18 J"/("1 g" * 1color(red)(cancel(color(black)(""^@"C")))))^(color(blue)("water's specific heat")) = "363.7 J g"^(-1)#

to increase its temperature by ${87}^{\circ} \text{C}$. Since your sample has a mass of $\text{26 g}$, it follows that you will need a total of

$26 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{g"))) * overbrace("363.7 J"/(1color(red)(cancel(color(black)("g")))))^(color(blue)("for a 87-"^@"C increase")) = color(darkgreen)(ul(color(black)("9,500 J}}}}$

The answer is rounded to two sig figs.