# A 36-gram sample of water has an initial temperature of 22°C. After the sample absorbs 1200 joules of heat energy, what is the final temperature of the sample?

Jan 11, 2017

${30.}^{\circ} \text{C}$

#### Explanation:

The key here is the specific heat of water, which you'll usually find list as

$c = {\text{4.18 J g"^(-1)""^@"C}}^{- 1}$

The specific heat of a substance 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, it takes $\text{4.18 J}$ of heat to increase the temperature of $\text{1 g}$ of water by ${1}^{\circ} \text{C}$.

Now, you know that your sample has a mass of $\text{36 g}$ and that you supply it with $\text{1200 J}$ of heat. Use the specific heat of water as a conversion factor to find the amount of heat needed to increase the temperature of this much water by ${1}^{\circ} \text{C}$

36 color(red)(cancel(color(black)("g"))) * "4.18 J"/(1color(red)(cancel(color(black)("g"))) * 1^@"C") = "150.5 J"""^@"C"^(-1)

This means that $\text{1200 J}$ of heat will increase the temperature of the sample by

1200 color(red)(cancel(color(black)("J"))) * (1^@"C")/(150.5color(red)(cancel(color(black)("J")))) = 7.973^@"C"

Therefore, the final temperature of the sample will be

T_"final" = 22^@"C" + 7.973^@"C" = color(darkgreen)(ul(color(black)(30.^@"C")))

I'll leave the answer rounded to two sig figs.