Question

Question #53ce8

Answer 1

`54.9^@"C"`

Explanation:

Your tool of choice here will be the equation

`color(blue)(ul(color(black)(q = m * c * DeltaT)))`

Here

• `q` is the heat absorbed by the water
• `m` is the mass of water
• `c` is the specific heat of liquid water, equal to `"4.184 J g"^(-1)""^@"C"^(-1)`
• `DeltaT` is the change in temperature, calculated as the difference between the final temperature and the initial temperature of the sample

Now, the idea here is that you need to figure out the initial temperature of the sample given the fact that you needed `5.19 * 10^5` `"J"` of heat to bring this sample to `100^@"C"`.

Mind you, you're going from liquid water at an initial temperature to liquid water at `100^@"C"`, i.e. you're not putting in any heat to convert the water to vapor at `100^@"C"`.

So if you take `T_i""^@"C"` to be the initial temperature of the water, you can say that you have

`DeltaT = (100 - T_i)^@"C"`

So now all you have to do is to use the equation to find the value of `T_i`.

`q = m * c * (100 - T_i)^@"C"`

Plug in your values to get

`5.19 * 10^5 color(red)(cancel(color(black)("J"))) = 2750color(red)(cancel(color(black)("g"))) * 4.184 color(red)(cancel(color(black)("J"))) color(red)(cancel(color(black)("g"^(-1)))) color(red)(cancel(color(black)(""^@"C"^(-1)))) * (100 - T_i) color(red)(cancel(color(black)(""^@"C")))`

Rearrange to get

`2750 * 4.184 * T_i = 2750 * 4.184 * 100 - 5.19 * 10^5`

You will end up with

`T_i = (2750 * 4.184 * 100 - 5.19 * 10^5)/(2750 * 4.184) = 54.9`

You can thus say that the initial temperature of the water was

`color(darkgreen)(ul(color(black)("initial temperature" = 54.9^@"C")))`

The answer is rounded to three sig figs.