Chemistry; Calorimetry; Calculating Heat Capacity; Help Please?!

One piece of copper jewelry at 105C has exactly twice the mass of another piece, which is at 45C. Both pieces are placed inside a calorimeter whose heat capacity is negligible. What is the final temperature inside the calorimeter?

1 Answer
Sep 8, 2017

Answer:

#85^@"C"#

Explanation:

The idea here is that the heat lost by the piece of jewelry that starts at a higher temperature will be equal to the heat gained by the piece of jewelry that starts at a lower temperature.

#color(blue)(ul(color(black)(q_"gained" = - q_"lost")))" " " "color(darkorange)("( * )")#

The minus sign is used here because the heat given off by the jewelry that cools off carries a negative sign!

Now, you can calculate the heat gained or given off by a substance by using the equation

#color(blue)(ul(color(black)(q = m * c * DeltaT)))#

Here

  • #m# is the mass of the sample
  • #c# is the specific heat of the substance
  • #DeltaT# is the change in temperature, calculated as the difference between the final temperature and the initial temperature of the substance

If you take #T_"f"# #""^@"C"# to be the final temperature inside the calorimeter, you can say that the heat gained by the piece that starts at #45^@"C"# is equal to

#q_"gained" = m * c * (T_"f" - 45)^@"C"#

Similarly, the heat lost by the piece that starts at #105^@"C"# will be equal to--keep in mind that this piece has twice the mass, or #2m#

#q_"lost" = 2m * c * (T_"f" - 105)^@"C"#

Use equation #color(darkorange)("( * )")# to say that--since both pieces of jewelry are made of copper, you know that they will have the same value for #c#

#color(red)(cancel(color(black)(m))) * color(red)(cancel(color(black)(c))) * (T_"f" - 45) color(red)(cancel(color(black)(""^@"C"))) = - 2color(red)(cancel(color(black)(m))) * color(red)(cancel(color(black)(c))) * (T_"f" - 105)color(red)(cancel(color(black)(""^@"C")))#

This is equivalent to

#T_"f" - 45 = -2T_"f" + 210#

Solve for #T_"f"# to find

#3T_"f" = 210 + 45#

#T_"f" = 255/3 = 85#

Therefore, you can say that the final temperature of the mixture is equal to

#color(darkgreen)(ul(color(black)("final temperature" = 85^@"C")))#

The answer is rounded to two sig figs.