# Question da3fc

Oct 27, 2017

$\text{18,800 J}$

#### Explanation:

You know that the specific heat of water is equal to ${\text{4.184 J g"^(-1)""^@"C}}^{- 1}$. This value tells you that in order to increase the temperature of $\text{1 g}$ of water by ${1}^{\circ} \text{C}$, you need to provide $\text{4.184 J}$ of heat.

Now, you know that your sample has a mass of $\text{51.4 g}$, so the first thing that you can do is figure out how much heat is required to heat $\text{51.4 g}$ of water.

51.4 color(red)(cancel(color(black)("g"))) * "4.184 J"/(1color(red)(cancel(color(black)("g"))) * 1^@"C") = "215.06 J"""^@"C"^(-1)#

This tells you that in order to increase the temperature of $\text{51.4 g}$ of water by ${1}^{\circ} \text{C}$, you need to provide $\text{215.06 J}$ of heat. In your case, you need to heat the water by

${92.7}^{\circ} \text{C" - 5.6^@"C" = 87.1^@"C}$

which means that you will need a total energy input of

$87.1 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{^@"C"))) * overbrace("215.06 J"/(1color(red)(cancel(color(black)(""^@"C")))))^(color(blue)("for 51.4 g of water")) = color(darkgreen)(ul(color(black)("18,800 J}}}}$

The answer is rounded to three sig figs.

SIDE NOTE Don't be confused by the fact that you have two sig figs for the initial temperature of the water.

Notice that you're using this value in a subtraction

${92.7}^{\circ} \text{C" - 5.6^@"C" = 87.1^@"C}$

The result of this operation must be rounded to one decimal place. In this case, ${87.1}^{\circ} \text{C}$ is rounded to one decimal place and has three sig figs, which is what you used in the subsequent multiplication.