Question #1dda3

1 Answer
Feb 12, 2015

I believe this is a heat capacity problem.

The equation for heat capacity is Q=cmT, where Q = the heat lost or gained, c = specific heat, m = mass in grams, and T = the temperature change, (#"T"_f-"T"_i#).

The density and volume are used to calculate the mass, most likely of water. Mass = density x volume.

In order to determine the final temperature, #"T"_f#, complete the following steps.

Rearrange the heat capacity equation to isolate T.

#"T = "Q/"mc"#

Now plug in your known values to get T, the change in temperature.

For example, suppose 300.0 calories were added to 100.0 grams of water at an initial temperature of #"22"^("o")"C"#. Water's specific heat, c, is #"1.0 cal/g·"^("o")"C"#.

#"T="Q/"mc"# = #"300.0 cal"/("100.0g·1.0 cal/g·"^("o")"C"# = #"3.0"^("o")"C"#

The initial temperature, #"T"_i"# was #"22"^("o")"C"#. Now add the change in temperature, T = #"3.0"^("o")"C"# to the initial temperature and that is the final temperature.

#"T"_f"# = #"T"_i"# + #"T"# = #"22"^("o")"C"# + #"3.0"^("o")"C = 25"^("o")"C"#