# Question #fcc68

Aug 18, 2017

About $\text{5 cal}$, or about $\text{20 J}$. (The more exact value was $\text{20.92 J}$, but the $\text{5 cal}$ is not rounded.)

The heat energy $q$ required to raise the temperature of $\text{1 g}$ of water by ${1}^{\circ} \text{C}$ is one calorie, $\text{1 cal}$. That is the definition of the specific heat capacity $c$ for water, ${c}_{w}$:

${c}_{w} = \text{1 cal/g"^@ "C}$

And thus, to raise the temperature of a gram of water by ${5}^{\circ} \text{C}$, we assume the specific heat capacity of water stays constant in this temperature range and write

$\textcolor{b l u e}{q} = \left(\text{1 cal")/(cancel"g"cancel(""^@ "C")) xx 5cancel(""^@ "C") xx cancel"1 g" ~~ color(blue)("5 cal}\right)$

This is from an actual equation in disguise...

$q = {m}_{w} {c}_{w} \Delta T$

where we assigned the mass of water as ${m}_{w} = \text{1 g}$ and the change in temperature as $\Delta T = {T}_{2} - {T}_{1}$ $=$ ${5}^{\circ} \text{C" - 0^@ "C}$ $=$ ${5}^{\circ} \text{C}$.

But we could have figured it out from the units if the equation does not come to mind, as you saw above.

And in case you want this in different units, note that in terms of joules, $\text{J}$, there are $\text{4.184 J}$ in one $\text{cal}$ at ${25}^{\circ} \text{C}$, so we get approximately:

$5 \cancel{\text{cal" xx "4.184 J"/cancel"cal" ~~ ul"20.92 J}}$

to one significant figure, as you have given, it would end up being $\textcolor{b l u e}{\text{20 J}}$. We could denote the digits past the last significant one as ${20.}_{92}$.

(It would appear that we round down in the tens place for the wrong reason, but since $0$ in the ones place (the digit past our last significant digit) is less than $5$, we do round down.)

Aug 18, 2017

$\text{20 J}$

#### Explanation:

The key here is the amount of heat needed to increase the temperature of $\text{1 g}$ of water by ${1}^{\circ} \text{C}$ $\to$ the specific heat of water.

$\textcolor{b l u e}{\underline{\textcolor{b l a c k}{{c}_{\text{water" = "4.18 J g"^(-1)""^@"C}}^{- 1}}}}$

So, the specific heat of a substance tells you the amount of heat needed in order to increase the temperature of $\text{1 g}$ of that substance by ${1}^{\circ} \text{C}$.

In the case of water, you need $\text{4.18 J}$ of heat to increase the temperature of a single gram of water by ${1}^{\circ} \text{C}$.

This implies that in order to increase the temperature of $\text{1 g}$ of water by ${5}^{\circ} \text{C}$, you need $5$ times as much heat as you'd need to increase the temperature of $\text{1 g}$ of water by ${1}^{\circ} \text{C}$.

$5 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{^@"C"))) * overbrace("4.18 J"/("1 g" * 1color(red)(cancel(color(black)(""^@"C")))))^(color(blue)("the specific heat of water")) = color(darkgreen)(ul(color(black)("20 J}}}}$

The answer must be rounded to one significant figure.