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Measuring Heat Flow
One technique we can use to measure the amount of heat involved in a les-grizzlys-catalans.orgical or physical process is known as calorimetry. Calorimetry is used to measure amounts of heat transferred to or from a substance. To do so, the heat is exchanged with a calibrated object (calorimeter). The change in temperature of the measuring part of the calorimeter is converted into the amount of heat (since the previous calibration was used to establish its heat capacity). The measurement of heat transfer using this approach requires the definition of a system (the substance or substances undergoing the les-grizzlys-catalans.orgical or physical change) and its surroundings (the other components of the measurement apparatus that serve to either provide heat to the system or absorb heat from the system). Knowledge of the heat capacity of the surroundings, and careful measurements of the masses of the system and surroundings and their temperatures before and after the process allows one to calculate the heat transferred as described in this section.
A calorimeter is a device used to measure the amount of heat involved in a les-grizzlys-catalans.orgical or physical process.
The thermal energy change accompanying a les-grizzlys-catalans.orgical reaction is responsible for the change in temperature that takes place in a calorimeter. If the reaction releases heat (qrxn calorimeter > 0) and its temperature increases. Conversely, if the reaction absorbs heat (qrxn > 0), then heat is transferred from the calorimeter to the system (qcalorimeter Note
The amount of heat absorbed or released by the calorimeter is equal in magnitude and opposite in sign to the amount of heat produced or consumed by the reaction.
Because ΔH is defined as the heat flow at constant pressure, measurements made using a constant-pressure calorimeter (a device used to measure enthalpy changes in les-grizzlys-catalans.orgical processes at constant pressure) give ΔH values directly. This device is particularly well suited to studying reactions carried out in solution at a constant atmospheric pressure. A “student” version, called a coffee-cup calorimeter (Figure (PageIndex2)), is often encountered in general les-grizzlys-catalans.orgistry laboratories. Commercial calorimeters operate on the same principle, but they can be used with smaller volumes of solution, have better thermal insulation, and can detect a change in temperature as small as several millionths of a degree (10−6°C).
Before we practice calorimetry problems involving les-grizzlys-catalans.orgical reactions, consider a simpler example that illustrates the core idea behind calorimetry. Suppose we initially have a high-temperature substance, such as a hot piece of metal (M), and a low-temperature substance, such as cool water (W). If we place the metal in the water, heat will flow from M to W. The temperature of M will decrease, and the temperature of W will increase, until the two substances have the same temperature—that is, when they reach thermal equilibrium (Figure (PageIndex4)). If this occurs in a calorimeter, ideally all of this heat transfer occurs between the two substances, with no heat gained or lost by either the calorimeter or the calorimeter’s surroundings. Under these ideal circumstances, the net heat change is zero:
This relationship can be rearranged to show that the heat gained by substance M is equal to the heat lost by substance W:
The magnitude of the heat (change) is therefore the same for both substances, and the negative sign merely shows that (q_substance; M) and (q_substance; W) are opposite in direction of heat flow (gain or loss) but does not indicate the arithmetic sign of either q value (that is determined by whether the matter in question gains or loses heat, per definition). In the specific situation described, qsubstance M is a negative value and qsubstance W is positive, since heat is transferred from M to W.
Heat Transfer between Substances at Different Temperatures
A 360-g piece of rebar (a steel rod used for reinforcing concrete) is dropped into 425 mL of water at 24.0 °C. The final temperature of the water was measured as 42.7 °C. Calculate the initial temperature of the piece of rebar. Assume the specific heat of steel is approximately the same as that for iron (Table T4), and that all heat transfer occurs between the rebar and the water (there is no heat exchange with the surroundings).
The temperature of the water increases from 24.0 °C to 42.7 °C, so the water absorbs heat. That heat came from the piece of rebar, which initially was at a higher temperature. Assuming that all heat transfer was between the rebar and the water, with no heat “lost” to the surroundings, then heat given off by rebar = −heat taken in by water, or:
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.Answer:
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.
Assuming perfect heat transfer, heat given off by metal = −heat taken in by water, or:
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).Answer
(c_metal= 0.13 ;J/g; °C)
< Delta H_rxn=q_rxn=-q_calorimater=-mC_s Delta T label(PageIndex5) >
When 5.03 g of solid potassium hydroxide are dissolved in 100.0 mL of distilled water in a coffee-cup calorimeter, the temperature of the liquid increases from 23.0°C to 34.7°C. The density of water in this temperature range averages 0.9969 g/cm3. What is ΔHsoln (in kilojoules per mole)? Assume that the calorimeter absorbs a negligible amount of heat and, because of the large volume of water, the specific heat of the solution is the same as the specific heat of pure water.
Given: mass of substance, volume of solvent, and initial and final temperatures
Asked for: ΔHsoln
Strategy:Calculate the mass of the solution from its volume and density and calculate the temperature change of the solution. Find the heat flow that accompanies the dissolution reaction by substituting the appropriate values into Equation (PageIndex1). Use the molar mass of KOH to calculate ΔHsoln.
A To calculate ΔHsoln, we must first determine the amount of heat released in the calorimetry experiment. The mass of the solution is
< left (100.0 ; mL; H2O ight ) left ( 0.9969 ; g/ cancelmL ight )+ 5.03 ; g ; KOH=104.72 ; g >
The temperature change is (34.7°C − 23.0°C) = +11.7°C.
B Because the solution is not very concentrated (approximately 0.9 M), we assume that the specific heat of the solution is the same as that of water. The heat flow that accompanies dissolution is thus
< q_calorimater=mC_s Delta T =left ( 104.72 ; cancelg ight ) left ( dfrac4.184 ; Jcancelgcdot cancel^oC ight )left ( 11.7 ; ^oC ight )=5130 ; J =5.13 ; lJ >
The temperature of the solution increased because heat was absorbed by the solution (q > 0). Where did this heat come from? It was released by KOH dissolving in water. From Equation (PageIndex1), we see that
ΔHrxn = −qcalorimeter = −5.13 kJ
This experiment tells us that dissolving 5.03 g of KOH in water is accompanied by the release of 5.13 kJ of energy. Because the temperature of the solution increased, the dissolution of KOH in water must be exothermic.
C The last step is to use the molar mass of KOH to calculate ΔHsoln—the heat released when dissolving 1 mol of KOH:
< Delta H_soln= left ( dfrac5.13 ; kJ5.03 ; cancelg ight )left ( dfrac56.11 ; cancelg1 ; mol ight )=-57.2 ; kJ/mol >
Constant-pressure calorimeters are not very well suited for studying reactions in which one or more of the reactants is a gas, such as a combustion reaction. The enthalpy changes that accompany combustion reactions are therefore measured using a constant-volume calorimeter, such as the bomb calorimeter (A device used to measure energy changes in les-grizzlys-catalans.orgical processes. shown sles-grizzlys-catalans.orgatically in Figure (PageIndex3)). The reactant is placed in a steel cup inside a steel vessel with a fixed volume (the “bomb”). The bomb is then sealed, filled with excess oxygen gas, and placed inside an insulated container that holds a known amount of water. Because combustion reactions are exothermic, the temperature of the bath and the calorimeter increases during combustion. If the heat capacity of the bomb and the mass of water are known, the heat released can be calculated.
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