Question #c1015

1 Answer
Sep 20, 2015

Answer:

#50^@"C"#

Explanation:

In order to solve this problem, you need to know water's specific heat, which expresses the amount of heat needed to raise the temperature of #"1 g"# of water by #1^@"C"#.

#c_"water" = 4.18"J"/("g" ^@"C")#

Now, the amount of heat is given to you in calories, which means that you're going to have to convert it to Joules

#1000color(red)(cancel(color(black)("cal"))) * "4.18400 J"/(1color(red)(cancel(color(black)("cal")))) = "4184.0 J"#

The equation that establishes a relationship between heat added/removed and increase/decrease in temperature looks like this

#q = m * c * DeltaT" "#, where

#q# - heat absorbed/removed;
#m# - the mass of the substance;
#c# - its specific heat;
#DeltaT# - the change in temperature, defined as the difference between the final temperature and the initial temperature.

Rearrange the equation to solve for #DeltaT#

#DeltaT = q/(m * c)#

#DeltaT = (4184.0color(red)(cancel(color(black)("J"))))/(100color(red)(cancel(color(black)("g"))) * 4.18color(red)(cancel(color(black)("J")))/(color(red)(cancel(color(black)("g"))) ^@"C")) = 10.0 ""^@"C"#

This means that the final temperature of the water will be

#DeltaT = T_"final" - T_"initial"#

#T_"final" = "DeltaT" + T_"initial" = 10""^@"C" + 40""^@"C" = color(green)(50""^@"C")#