# 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?

##### 1 Answer

#### Answer:

#### 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

This tells you that the first thing to do here is to calculate the *change in temperature* that occurs when you provide

Your tool of choice here will this equation

#color(blue)(bar(ul(|color(white)(a/a)q = m * c * DeltaTcolor(white)(a/a)|)))" "# , where

*change in temperature*, defined as the difference between the **final temperature** and the **initial temperature**

Water has a **specific heat** of

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

Rearrange the equation to solve for

#q = m * c * DeltaT implies DeltaT = q/(m * 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 **higher** than the initial temperature of the sample.

You can thus say that you have

#DeltaT = T_"f" - T_"i"#

and

#T_"i" = T_"f" - DeltaT#

#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.