# How much heat (kJ) is needed to raise the temperature of 100.0 grams of water from 25.0°C to 50.0°C?

Jan 25, 2016

$\text{10.5 kJ}$

#### Explanation:

In order to be able to solve this problem, you will need to know the value of water's specific heat, which is listed as

c = 4.18"J"/("g" ""^@"C")

Now, let's assume that you don't know the equation that allows you to plug in your values and find how much heat would be needed to heat that much water by that many degrees Celsius.

Take a look at the specific heat of water. As you know, 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 water's case, you need to provide $\text{4.18 J}$ of heat per gram of water to increase its temperature by ${1}^{\circ} \text{C}$.

What if you wanted to increase the temperature of $\text{1 g}$ of water by ${2}^{\circ} \text{C}$ ? You'd need to provide it with

$\overbrace{\text{4.18 J")^(color(purple)("increase by 1"""^@"C")) + overbrace("4.18 J")^(color(purple)("increase by 1"""^@"C")) = overbrace(2 xx "4.18 J")^(color(green)("increase by 2"""^@"C}}$

To increase the temperature of $\text{1 g}$ of water by ${n}^{\circ} \text{C}$, you'd need to supply it with

$\overbrace{\text{4.18 J")^(color(purple)("increase by 1"""^@"C")) + overbrace("4.18 J")^(color(purple)("increase by 1"""^@"C")) + " ... " = overbrace(n xx "4.18 J")^(color(green)("increase by n"""^@"C}}$

Now let's say that you wanted to cause a ${1}^{\circ} \text{C}$ increase in a $\text{2-g}$ sample of water. You'd need to provide it with

overbrace("4.18 J")^(color(brown)("for 1 g of water")) + overbrace("4.18 J")^(color(brown)("for 1 g of water")) = overbrace(2 xx "4.18 J")^(color(green)("for 2 g of water"))

To cause a ${1}^{\circ} \text{C}$ increase in the temperature of $m$ grams of water, you'd need to supply it with

overbrace("4.18 J")^(color(brown)("for 1 g of water")) + overbrace("4.18 J")^(color(brown)("for 1 g of water")) + " ,,, " = overbrace(m xx "4.18 J")^(color(green)("for m g of water"))

This means that in order to increase the temperature of $m$ grams of water by ${n}^{\circ} \text{C}$, you need to provide it with

$\text{heat" = m xx n xx "specific heat}$

This will account for increasing the temperature of the first gram of the sample by ${n}^{\circ} \text{C}$, of the the second gram by ${n}^{\circ} \text{C}$, of the third gram by ${n}^{\circ} \text{C}$, and so on until you reach $m$ grams of water.

And there you have it. The equation that describes all this will thus be

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

$q$ - heat absorbed
$m$ - 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

In your case, you will have

$q = 100.0 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{g"))) * 4.18"J"/(color(red)(cancel(color(black)("g"))) color(red)(cancel(color(black)(""^@"C")))) * (50.0 - 25.0)color(red)(cancel(color(black)(""^@"C}}}}$

$q = \text{10,450 J}$

Rounded to three sig figs and expressed in kilojoules, the answer will be

"10,450" color(red)(cancel(color(black)("J"))) * "1 kJ"/(10^3color(red)(cancel(color(black)("J")))) = color(green)("10.5 kJ")