Question

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?

Answer 1

`42.9^@"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^@"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 * 10^5"J"` of heat to `"2290 g"` of water.

Your tool of choice here will this equation

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

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

Water has a specific heat of `"4.18 J g"^(-1)""^@"C"^(-1)`

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

Rearrange the equation to solve for `DeltaT`

`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 `100^@"C"` and that this final temperature is `57.14^@"C"` 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.