# 5.4 Thermochemistry: calorimetry  (Page 3/14)

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$\left(0.449\phantom{\rule{0.2em}{0ex}}\text{J/g °C}\right)\left(360\phantom{\rule{0.2em}{0ex}}\text{g}\right)\left(42.7\phantom{\rule{0.2em}{0ex}}\text{°C}-{T}_{\text{i,rebar}}\right)=\left(4.184\phantom{\rule{0.2em}{0ex}}\text{J/g °C}\right)\left(425\text{g}\right)\left(42.7\phantom{\rule{0.2em}{0ex}}\text{°C}-24.0\phantom{\rule{0.2em}{0ex}}\text{°C}\right)$
${T}_{\text{i,rebar}}=\phantom{\rule{0.2em}{0ex}}\frac{\left(4.184\phantom{\rule{0.2em}{0ex}}\text{J/g °C}\right)\left(425\phantom{\rule{0.2em}{0ex}}\text{g}\right)\left(42.7\phantom{\rule{0.2em}{0ex}}\text{°C}-24.0\phantom{\rule{0.2em}{0ex}}\text{°C}\right)}{\left(0.449\phantom{\rule{0.2em}{0ex}}\text{J/g °C}\right)\left(360\phantom{\rule{0.2em}{0ex}}\text{g}\right)}\phantom{\rule{0.2em}{0ex}}+42.7\phantom{\rule{0.2em}{0ex}}\text{°C}$

Solving this gives T i,rebar = 248 °C, so the initial temperature of the rebar was 248 °C.

A 248-g piece of copper is dropped into 390 mL of water at 22.6 °C. The final temperature of the water was measured as 39.9 °C. Calculate the initial temperature of the piece of copper. Assume that all heat transfer occurs between the copper and the water.

The initial temperature of the copper was 335.6 °C.

A 248-g piece of copper initially at 314 °C is dropped into 390 mL of water initially at 22.6 °C. Assuming that all heat transfer occurs between the copper and the water, calculate the final temperature.

The final temperature (reached by both copper and water) is 38.8 °C.

This method can also be used to determine other quantities, such as the specific heat of an unknown metal.

## Identifying a metal by measuring specific heat

A 59.7 g piece of metal that had been submerged in boiling water was quickly transferred into 60.0 mL of water initially at 22.0 °C. The final temperature is 28.5 °C. Use these data to determine the specific heat of the metal. Use this result to identify the metal.

## Solution

Assuming perfect heat transfer, heat given off by metal = −heat taken in by water , or:

${q}_{\text{metal}}=\text{−}{q}_{\text{water}}$

In expanded form, this is:

${c}_{\text{metal}}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{m}_{\text{metal}}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\left({T}_{\text{f,metal}}-{T}_{\text{i, metal}}\right)=\text{−}{c}_{\text{water}}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{m}_{\text{water}}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\left({T}_{\text{f,water}}-{T}_{\text{i,water}}\right)$

Noting that since the metal was submerged in boiling water, its initial temperature was 100.0 °C; and that for water, 60.0 mL = 60.0 g; we have:

$\left({c}_{\text{metal}}\right)\left(59.7\phantom{\rule{0.2em}{0ex}}\text{g}\right)\left(28.5\phantom{\rule{0.2em}{0ex}}\text{°C}-100.0\phantom{\rule{0.2em}{0ex}}\text{°C}\right)=-\left(4.18\phantom{\rule{0.2em}{0ex}}\text{J/g °C}\right)\left(60.0\phantom{\rule{0.2em}{0ex}}\text{g}\right)\left(28.5\phantom{\rule{0.2em}{0ex}}\text{°C}-22.0\phantom{\rule{0.2em}{0ex}}\text{°C}\right)$

Solving this:

${c}_{\text{metal}}=\phantom{\rule{0.2em}{0ex}}\frac{-\left(4.184\phantom{\rule{0.2em}{0ex}}\text{J/g °C}\right)\left(60.0\phantom{\rule{0.2em}{0ex}}\text{g}\right)\left(6.5\phantom{\rule{0.2em}{0ex}}\text{°C}\right)}{\left(59.7\phantom{\rule{0.2em}{0ex}}\text{g}\right)\left(-71.5\phantom{\rule{0.2em}{0ex}}\text{°C}\right)}\phantom{\rule{0.2em}{0ex}}=0.38\phantom{\rule{0.2em}{0ex}}\text{J/g °C}$

Comparing this with values in [link] , our experimental specific heat is closest to the value for copper (0.39 J/g °C), so we identify the metal as copper.

A 92.9-g piece of a silver/gray metal is heated to 178.0 °C, and then quickly transferred into 75.0 mL of water initially at 24.0 °C. After 5 minutes, both the metal and the water have reached the same temperature: 29.7 °C. Determine the specific heat and the identity of the metal. (Note: You should find that the specific heat is close to that of two different metals. Explain how you can confidently determine the identity of the metal).

c metal = 0.13 J/g °C

This specific heat is close to that of either gold or lead. It would be difficult to determine which metal this was based solely on the numerical values. However, the observation that the metal is silver/gray in addition to the value for the specific heat indicates that the metal is lead.

When we use calorimetry to determine the heat involved in a chemical reaction, the same principles we have been discussing apply. The amount of heat absorbed by the calorimeter is often small enough that we can neglect it (though not for highly accurate measurements, as discussed later), and the calorimeter minimizes energy exchange with the surroundings. Because energy is neither created nor destroyed during a chemical reaction, there is no overall energy change during the reaction. The heat produced or consumed in the reaction (the “system”), q reaction , plus the heat absorbed or lost by the solution (the “surroundings”), q solution , must add up to zero:

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