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
Sep 15, 2016

Answer:

#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.