# Question 71e95

Apr 3, 2016

${58.2}^{\circ} \text{C}$

#### Explanation:

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

$\textcolor{p u r p \le}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{{c}_{\text{water" = "4.18 J g"^(-1)""^@"C}}^{- 1}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

A substance's specific heat tells you how much heat is needed in order to increase the temperature of $\text{1 g}$ of that substance by ${1}^{\circ} \text{C}$.

In your case, you can say that in order to increase the temperature of $\text{1 g}$ of water by ${1}^{\circ} \text{C}$, you need to provide it with $\text{4.18 J}$ of heat.

Keep in mind that this much heat must be applied per gram, per degree Celsius.

Now, you know that your sample of water has a mass of $\text{150. g}$, and that it absorbs a total of $\text{21.8 kJ}$ of heat. Convert the heat from kilojoules to joules by using the conversion factor

$\text{1 kJ" = 10^3"J}$

to get

21.8color(red)(cancel(color(black)("kJ"))) * (10^3"J")/(1color(red)(cancel(color(black)("kJ")))) = "21,800 J"

At this point, you can determine how much heat would be needed in order to increase the temperature of this sample of water by ${1}^{\circ} \text{C}$. Well, you know that you need $\text{4.18 J}$ per gram in order to get a ${1}^{\circ} \text{C}$ increase in temperature, so $\text{150. g}$ of water will require

150. color(red)(cancel(color(black)("g"))) * overbrace("4.18 J"/(1color(red)(cancel(color(black)("g")))))^(color(purple)("heat needed for 1"^@"C increase for 1 g")) = "627 J"

If you need $\text{627 J}$ of heat in order to increase the temperature of $\text{150 g}$ of water by ${1}^{\circ} \text{C}$, it follows that the difference between this much heat and the $\text{21,800 J}$ you supplied went in increasing the sample's temperature.

You can thus say that

$\text{21,800"color(red)(cancel(color(black)("J"))) * overbrace((1^@"C")/(627color(red)(cancel(color(black)("J")))))^(color(blue)("heat needed for 1"^@"C for 150. g")) = 34.77^@"C}$

If the sample started at an initial temperature of ${23.4}^{\circ} \text{C}$, you can say that its final temperature will be

T_"final" = 23.4^@"C" + 34.77^@"C" = color(green)(|bar(ul(color(white)(a/a)58.2^@"C"color(white)(a/a)|)))

The answer is rounded to three sig figs.

ALTERNATIVE APPROACH

You can get the same result by suing the equation

$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} q = m \cdot c \cdot \Delta T \textcolor{w h i t e}{\frac{a}{a}} |}}} \text{ }$, where

$q$ - the amount of heat gained / lost
$m$ - the mass of the sample
$c$ - the specific heat of the substance
$\Delta T$ - the change in temperature, defined as the difference between the final temperature and the initial temperature

$\Delta T = \frac{q}{m \cdot c}$
DeltaT = ("21,800" color(red)(cancel(color(black)("J"))))/(150. color(red)(cancel(color(black)("g"))) * 4.18color(red)(cancel(color(black)("J")))color(red)(cancel(color(black)("g"^(-1))))""^@"C"^(-1)) = 34.77^@"C"
T_"final" = 23.4^@"C" + DeltaT = color(green)(|bar(ul(color(white)(a/a)58.2^@"C"color(white)(a/a)|)))#