# How many calories are absorbed when 500 grams of water (1 cal/g c) is heated from 50 to 100 degrees Celsius?

May 28, 2017

$\text{30,000 cal}$

#### Explanation:

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

In your case, water is said to have a specific heat equal to

${c}_{\text{water" = "1 cal g"^(-1)""^@"C}}^{- 1}$

This means that in order to increase the temperature of $\text{1 g}$ of water by ${1}^{\circ} \text{C}$, you need to provide it with $\text{1 cal}$ of heat.

So, how much heat would be needed to increase the temperature of $\text{500 g}$ of water by ${1}^{\circ} \text{C}$ ?

500 color(red)(cancel(color(black)("g"))) * overbrace("1 cal"/(1color(red)(cancel(color(black)("g")))))^(color(blue)("for 1"^@"C")) = "500 cal"

This means that in order to increase the temperature of $\text{500 g}$ of water by

${100}^{\circ} \text{C" - 50^@"C" = 50^@"C}$

you need to provide it with

$50 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{^@"C"))) * overbrace("500 cal"/(1color(red)(cancel(color(black)(""^@"C")))))^(color(blue)("for 500 g")) = "25,000 cal" = color(darkgreen)(ul(color(black)("30,000 cal}}}}$

The answer must be rounded to one significant figure, the number of sig figs you have for your values.

Keep in mind that this much heat is needed in order to get $\text{500 g}$ of water from liquid at ${50}^{\circ} \text{C}$ to liquid at ${100}^{\circ} \text{C}$.