# Water (2290 g) is heated until it just begins to boil. If the water absorbs 5.47xx10^5 J of heat in the process what was the initial temperature of the water?

Sep 15, 2016

#### Answer:

${42.9}^{\circ} \text{C}$

#### Explanation:

The idea here is that the problem is providing you with the amount of heat needed to raise the temperature of a given sample of water from an initial temperature to its boiling point, i.e. to ${100}^{\circ} \text{C}$.

This tells you that the first thing to do here is to calculate the change in temperature that occurs when you provide $5.47 \cdot {10}^{5} \text{J}$ of heat to $\text{2290 g}$ of water.

Your tool of choice here will this equation

$\textcolor{b l u e}{\overline{\underline{| \textcolor{w h i t e}{\frac{a}{a}} q = m \cdot c \cdot \Delta T \textcolor{w h i t e}{\frac{a}{a}} |}}} \text{ }$, where

$q$ - the amount of heat gained
$m$ - the mass of the sample
$c$ - the specific heat of the substance
$\Delta T$ - the change in temperature, defined as the difference between the final temperature and the initial temperature

Water has a specific heat of ${\text{4.18 J g"^(-1)""^@"C}}^{- 1}$

http://www.engineeringtoolbox.com/water-thermal-properties-d_162.html

Rearrange the equation to solve for $\Delta T$

$q = m \cdot c \cdot \Delta T \implies \Delta T = \frac{q}{m \cdot c}$

Plug in your values to find

DeltaT = (5.47 * 10^5 color(red)(cancel(color(black)("J"))))/(2290 color(red)(cancel(color(black)("g"))) * 4.18 color(red)(cancel(color(black)("J"))) color(red)(cancel(color(black)("g"^(-1))))""^@"C"^(-1)) = 57.14^@"C"

So, you know that the final temperature of the water is ${100}^{\circ} \text{C}$ and that this final temperature is ${57.14}^{\circ} \text{C}$ higher than the initial temperature of the sample.

You can thus say that you have

$\Delta T = {T}_{\text{f" - T_"i}}$

and

${T}_{\text{i" = T_"f}} - \Delta T$

T_"i" = 100^@"C" - 57.14^@"C" = color(green)(bar(ul(|color(white)(a/a)color(black)(42.9^@"C")color(white)(a/a)|)))

I'll leave the answer rounded to three sig figs and one decimal place.