# Question #53ce8

##### 1 Answer

#### 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 themassof water#c# is thespecific heatof liquid water, equal to#"4.184 J g"^(-1)""^@"C"^(-1)# #DeltaT# is thechange 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

Mind you, you're going from **liquid water** at an initial temperature to **liquid water** at

So if you take

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

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

#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**.