# Question d5fe2

Sep 16, 2015

You need $\text{10,000 J}$ worth of heat.

#### Explanation:

Start by taking a look at sand's specific heat, which is said to be equal to 664"J"/("kg" ^@"C").

When given in Joules per kilogram Kelvin, the specific heat of a substance tells you how much heat is required to increase the temperature of $\text{1 kg}$ of that substance by ${1}^{\circ} \text{C}$.

In your case, if you add $\text{664 J}$ to $\text{1 kg}$ of sand, you will increase its temperature by ${1}^{\circ} \text{C}$.

Since you need to increase its temperature by ${20}^{\circ} \text{C}$, it follows that you must supply 20 times more heat than you would have supplied to increase the temperature by only ${1}^{\circ} \text{C}$.

The equation that links added/removed heat to increase/decrease in temperature looks like this

$q = m \cdot c \cdot \Delta T \text{ }$, where

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

So, use the value given to yout to find

$q = 1 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{kg"))) * 664"J"/(color(red)(cancel(color(black)("kg"))) * color(red)(cancel(color(black)(""^@"C")))) * (50-30)color(red)(cancel(color(black)(""^@"C}}}}$

q = "13,280 J" = color(green)("10,000 J")#

The answer is rounded to one sig fig, the number of sig figs you gave for the mass of the sample.