# Question 3c48f

Dec 30, 2016

Here's what I got.

#### Explanation:

The key here is the specific heat of water, which is usually given as

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

The specific heat of a substance is the amount of heat that must be provided in order to increase the temperature of $\text{1 g}$ of that substance by ${1}^{\circ} \text{C}$.

In your case, you need to provide $\text{4.18 J}$ of heat in order to heat $\text{1 g}$ of water by ${1}^{\circ} \text{C}$.

Now, the problem provides you the amount of heat that you have at your disposal and the desired change in temperature, which is equal to

$\Delta T = {75.0}^{\circ} \text{C" - 20.0^@"C" = 55.0^@"C}$

Use the specific heat of water to calculate how much heat would be needed in order to increase the temperature of $\text{1 g}$ of water by ${55}^{\circ} \text{C}$

55.0 color(red)(cancel(color(black)(""^@"C"))) * "4.18 J"/("1 g" * 1color(red)(cancel(color(black)(""^@"C")))) = "229.9 J g"^(-1)#

So, you must add $\text{229.9 J}$ for every $\text{1 g}$ of water in order to cause its temperature to increase by ${55.0}^{\circ} \text{C}$. This means that $\text{8000 J}$ of heat will get you

$8000 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{J"))) * overbrace("1 g"/(229.9 color(red)(cancel(color(black)("J")))))^(color(blue)("needed for 55" ""^@"C increase")) = color(darkgreen)(ul(color(black)("35.8 g}}}}$

I'll leave the answer rounded to three sig figs, but keep in mind that you only provided one significant figure for the amount of heat you have at your disposal.