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

##### 1 Answer

#### 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

In water's case, you need to provide **per gram** of water to increase its temperature by

What if you wanted to increase the temperature of

#overbrace("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

#overbrace("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

#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

#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

#"heat" = m xx n xx "specific heat"#

This will account for increasing the temperature of the *first gram* of the sample by *second gram* by *third gram* by

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

#color(blue)(q = m * c * DeltaT)" "# , where

*final temperature* minus *initial temperature*

In your case, you will have

#q = 100.0 color(red)(cancel(color(black)("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 = "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")#