# The temperature of a sample of water changes from 10° C to 20°C when the sample absorbs 418 joules of heat. What is the mass of the sample?

Dec 15, 2015

$\text{10 g}$

#### Explanation:

Right from the start, just by inspecting the values given, you can say that the answer will be $\text{10 g}$.

Now, here's what that is the case.

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

Water has a specific heat of approximately 4.18"J"/("g" ""^@"C"). This tells you that in order to increase the temperature of $\text{1 g}$ of water by ${1}^{\circ} \text{C}$, you need to provide $\text{4.18 J}$ of heat.

Now, how much heat would be required to increase the temperature of $\text{1 g}$ of water by ${10}^{\circ} \text{C}$?

Well, you'd need $\text{4.18 J}$ to increase it by ${1}^{\circ} \text{C}$, another $\text{4.18 J}$ to increase it by another ${1}^{\circ} \text{C}$, and so on. This means that you'd need

$\text{4.18 J" xx 10 = "41.8 J}$

to increase the temperature of $\text{1 g}$ of water by ${10}^{\circ} \text{C}$.

Now look at the value given to you. If you need $\text{41.8 J}$ to increase the temperature of $\text{1 g}$ of water by ${10}^{\circ} \text{C}$, what mass of water would require $10$ times as much heat to increase its temperature by ${10}^{\circ} \text{C}$?

$\text{1 g" xx 10 = "10 g}$

Mathematically, you can calculate this by using the equation

$\textcolor{b l u e}{q = m \cdot c \cdot \Delta T} \text{ }$, where

$q$ - heat absorbed/lost
$m$ - the mass of the sample
$c$ - the specific heat of the substance
$\Delta T$ - the change in temperature, defined as final temperature minus initial temperature

Plug in your values to get

$418 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{J"))) = m * 4.18color(red)(cancel(color(black)("J")))/("g" color(red)(cancel(color(black)(""^@"C")))) * (20 - 10)color(red)(cancel(color(black)(""^@"C}}}}$

$m = \frac{418}{4.18 \cdot 10} = \text{10 g}$