2.53)
29.5x - 6162.55 = 16414.89 - 54.3x
83.8x = 22577.44
x = 269.4 K
In case, you're not sure what happened to the 2.53, I simply divided both sides by 2.53 first.
Example #5: A sheet of nickel weighing 10.0 g and at a temperature of 18.0 °C is placed flat on a sheet of iron weighing 20.0 g and at a temperature of 55.6 °C. What is the final temperature of the combined metals? Assume that no heat is lost to the surroundings.
Solution:
This problem requires us to find the specific heats for nickel and iron. To do this, we will use this site. The values given are respectively, 0.54 J g¯1 °C¯1 and 0.46 J g¯1 °C¯1
Notice that the units on the site are kJ kg¯1 K¯1. In addition, notice that I wrote J g¯1 °C¯1. Also, notice that there is no numerical difference when using either specific heat unit (the kJ one or the J one). In other words:
one kJ kg¯1 K¯1 = one J g¯1 °C¯1
The left-hand unit is the IUPAC-approved one; the one on the right-hand side is the one in most common use.
On to the solution:
qlost = qgain
Therefore:
(20.0) (55.6 - x) (0.46) = (10.0) (x - 18.0) (0.54)
9.2 (55.6 - x) = 5.4 (x - 18)
511.52 - 9.2x = 5.4x - 97.2
14.6x = 608.72
x = 41.7 °C
