# Question d4a35

Apr 14, 2016

$\text{12.8 kJ}$

#### Explanation:

The key to this problem lies with the value of water's specific heat, which is said to be equal to

${c}_{w} = {\text{4.184 J g"^(-1)""^@"C}}^{- 1}$

Now, 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, water's specific heat tells you that you need $\text{4.184 J}$ of energy in the form of heat in order to increase the temperature of $\text{1 g}$ of water by ${1}^{\circ} \text{C}$.

Notice that specific heat is expressed in joules per gram per degree Celsius. This means that you can break up the calculation into two parts

• determine how much heat you need to add in order to increase the temperature of $\text{45.0 g}$ of water by ${1}^{\circ} \text{C}$

• use this value to determine how much heat would be needed in order to increase the sample's temperature by $\Delta T$

In this case, $\Delta T$, which represents change in temperature, will be equal to

$\Delta T = {T}_{\text{final" - T_"initial}}$

$\Delta T = {74.0}^{\circ} \text{C" - 6.2^@"C" = 67.8^@"C}$

So, to increase the temperature of $\text{45.0 g}$ of water by ${1}^{\circ} \text{C}$, you'd need

45.0 color(red)(cancel(color(black)("g"))) * overbrace(("4.184 J"^@"C"^(-1))/(1color(red)(cancel(color(black)("g")))))^(color(purple)("heat needed per 1"^@"C")) = "188.28 J"^@"C"^(-1)

If this is how much heat you need to produce a ${1}^{\circ} \text{C}$ increase in temperature for a $\text{45.0-g}$ sample of water, it follows that a ${67.8}^{\circ} \text{C}$ increase in temperature would require

67.8 color(red)(cancel(color(black)(""^@"C"))) * "188.28 J"/(1color(red)(cancel(color(black)(""^@"C")))) = "12,765.4 J"

I'll leave the answer rounded to three sig figs and express it in kilojoules

"heat needed" = color(green)(|bar(ul(color(white)(a/a)"12.8 kJ"color(white)(a/a)|)))#

ALTERNATIVE APPROACH

You can also solve this problem by using the following 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

Plug in your values to get

$q = 45.0 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{g"))) * "4.184 J" color(red)(cancel(color(black)("g"^(-1)))) color(red)(cancel(color(black)(""^@"C"^(-1)))) * 67.8color(red)(cancel(color(black)(""^@"C}}}}$

$q = \textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \text{12.8 kJ} \textcolor{w h i t e}{\frac{a}{a}} |}}}$