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

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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?

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

#### Answer:

#### 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 signis used here because the heatgiven offby the jewelry thatcools offcarries 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 themassof the sample#c# is thespecific heatof the substance#DeltaT# is thechange in temperature, calculated as the difference between thefinal temperatureand theinitial temperatureof the substance

If you take **final temperature** inside the calorimeter, you can say that the heat **gained** by the piece that starts at

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

Similarly, the heat **lost** by the piece that starts at **twice the mass**, or

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

Use equation

#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

#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**.