Specific Heat Capacity

PRE-LAB ASSIGNMENTS:

To be assigned by your lab instructor.

STUDENT LEARNING OUTCOMES:

The First Law of Thermodynamics Thermal energy as a function of temperature The units of energy Sign conventions in heat flow Heat capacity Specific heat capacity Factors in experimental error

EXPERIMENTAL GOALS:

You will measure the specific heat capacity of a known metal. Then you will measure the specific heat capacity of an unknown to identify the metal.

INTRODUCTION:

Heat Capacity

A large part of chemistry is studying and understanding the physical and chemical changes that matter undergoes. The driving force for most chemical and physical changes is energy, often in the form of thermal energy (heat). Temperature and heat are not the same quantity. The temperature of an object is an indicator of the amount of thermal energy it “possesses”, but heat is a measure of the transfer of energy to or from an object. We know that the more a substance is heated, the more its temperature will rise. This relationship is a direct proportionality that may be expressed as , where q is the heat absorbed or released by the system and ΔT is its temperature change (ΔT =Tfinal – Tinitial). This proportionality may be also be written as an equality,

or [1]

The constant of proportionality is the heat capacity, C, and is defined as the quantity of heat required to change an object’s temperature by one unit of temperature (usually 1 oC or 1 K). Hence we can rewrite Equation 1 as

[2]

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It follows that heat capacity has units related to and is commonly expressed as

(also written as J/°C or J °C-1).

The heat capacity of a system is dependent on a few factors. The heat capacity is dependent on the amount of a sample present as shown in Figure 1, and is thus an extensive property: a larger amount of sample requires more heat than a smaller amount of the same substance to raise the substance’s temperature by one unit. In addition to the amount of substance present, the kind of substance has a direct bearing on its heat capacity. The same amounts of different substances usually have different heat capacities, and heat capacity is often used for samples of a fixed size. It would be more convenient if the heat capacity depended only on the type of material, and not the amount of material. If heat capacity were an intensive property, independent of the amount of substance, it could be used as an analytical tool to identify a substance, much like density.

Figure 1. Heat required for 1 oC change in temperature showing dependenceof heat capacity, C, on amount of substance.

Specific Heat Capacity

In order to overcome the fact that heap capacity varies with the amount of a substance, chemists often use the specific heat capacity, Cs, also known as the specific heat. The specific heat capacity of a substance is defined as the amount of heat required to raise the temperature of 1 gram of the substance by 1oC (see Figure 2).

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100 g 200 g

C = q C' = 2q

Figure 2. Specific heat capacity.

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The relationship between the thermal energy and specific heat capacity may be expressed as

[3]

Equation 3 is similar to Equation 2 with the additional parameter of mass, m. The heat capacity, C, of a given sample may be converted to its specific heat capacity, Cs, by dividing the mass of the sample into C to give

By accounting for the amount of substance, the specific heat capacity depends only on the kind of substance and is therefore an intensive property. This quantity has units related to

. The units of specific heat are commonly expressed as (also written as

J/g·°C or J g-1 °C-1).

In this laboratory, a calorimeter is used to measure the specific heat of various known metals. A calorimeter is an apparatus used to measure the heat released or absorbed by a physical or chemical process. The calorimeter used here is composed of a pair of stacked Styrofoam coffee-cups with a lid (see Figure 3 in the Procedure section). The lid has two holes, one for a thermometer and the other for a stirring wire, and helps minimize heat loss to the atmosphere. In addition to measuring the specific heat of a known metal, an unknown metal will be identified based on the determination of its specific heat capacity.

The specific heat can be determined by immersing a known mass of warm metal of a known temperature into a known volume of water, at a known, lower temperature. Heat energy will flow from the hot metal to the water until, after a period of time, the water and metal achieve the same temperature. From the First Law of Thermodynamics, the sum of the heat lost by the metal and the heat gained by the water in a perfect experiment is qmetal + qwater = 0, which we can rearrange as

qmetal = -qwater [4]

The negative sign in Equation 4 comes about because of the ΔT term (see Equation 3). The final temperature of the metal is lower than the initial temperature, and the resulting value of qmetal will be negative. By convention, the flow of heat out of a body is a negative quantity.

An ideal calorimeter would provide perfect insulation so that its contents, the system, could undergo its temperature change without any loss or gain of heat to the surroundings or to the calorimeter itself. Such a calorimeter does not exist. Instead, we must build an apparatus that gains (or loses) a minimum amount of heat energy itself while also minimizing the flow of heat into or out of the Universe. Experimentally, the total heat gained by the water that we measure will not be the actual heat lost by the metal since the apparatus will gain some of the heat as well as permit some leakage to the atmosphere. In this experiment, the heat lost by leakage to the atmosphere will be ignored, since the Styrofoam calorimeter is a reasonably good

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insulator. However, the heat lost by the system to the calorimeter is small, but not negligible, and it must be considered. With this correction, Equation 4 becomes

qmetal = -(qwater + qcal) [5]

To determine calorimeter, the heat capacity of the calorimeter, Ccal, must first be determined. With Ccal, the heat absorbed by the calorimeter may be accounted for in the determination of the specific heat of a metal. The determination of Ccal is called the calibration of the calorimeter.

To calibrate the calorimeter, warm water of a known mass and temperature is mixed with a known mass and temperature of cooler water in the apparatus. In an ideal apparatus, the heat gained by the cooler water would exactly equal the heat lost by the warmer water. However, in the real world, the heat lost by the warm water will be equal to the heat gained by the cooler water plus heat lost to the calorimeter,

qlost by hot = -(qgained by cold + qcal) [6]

Rearrangement yields

qcal = -(qlost by hot + qgained by cold) [7]

which can be used to determine the heat capacity of the calorimeter.

In this experiment, the actual calibration of the calorimeter will not be performed. Instead the Ccal is given as 8.36 J/oC. With this value of Ccal for the coffee-cup calorimeter, Equations 3 and 5 may be used to determine the specific heat of a metal. Table 1 lists the specific heat capacities of various metals.

Table 1. Specific heat capacities of various metals.

Metal

Specific HeatCapacity, Cs

(J/g∙oC) Metal

Specific HeatCapacity, Cs

(J/g∙oC)Lead 0.129 Cobalt 0.421Gold 0.129 Nickel 0.444Tungsten 0.132 Iron 0.449Platinum 0.133 Chromium 0.450Neodymium 0.191 Manganese 0.479Tin 0.228 Titanium 0.522Silver 0.235 Potassium 0.757Strontium 0.306 Aluminum 0.897Copper 0.385 Magnesium 1.02Zinc 0.388 Sodium 1.23

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SAMPLE CALCULATION OF THE CALIBRATION OF A CALORIMETER:

Water with a mass of 50.0 g at 80.1 oC was added to 50.0 g of water at 27.2 oC. Upon mixing, the final temperature attained was 51.9 oC. What is the heat capacity of the calorimeter, Ccal? (Note: This example calculates the heat capacity of the calorimeter, whereas in the laboratory you will need to calculate the specific heat of the metal. The calculations are similar, but the distinction needs to be recognized.)

Heat lost by warm water:qlost = (50.0 g)∙(4.18 J/g∙oC)∙(51.9 oC – 80.1 oC) = -5890 J

Heat gained by cool water:qgained = (50.0 g)∙(4.18 J/g∙oC)∙(51.9 oC – 27.2 oC) = 5160 J

Heat absorbed by calorimeter:qcal = -(-5890 J + 5160 J) = 730 J

Heat capacity of calorimeter:

29 J/oC

PROCEDURE:

A. Measure the Cs of a Known Sample.

1. Obtain two nested coffee-cups. Zero the balance and weigh the empty, dry cups to the nearest milligram. Add approximately 50 mL of water to the cups and determine the mass of water by difference. Assemble the coffee-cup calorimeter as shown in Figure 3. First, insert the thermometer into the split stopper, and slide the stopper toward the top of the thermometer. Next, slide the thermometer into the center hole of the calorimeter lid. Place the ring stand clamp on the stopper, and adjust the clamp on the ring stand to where it will hold the thermometer bulb about 1 cm from the bottom of the cup. Finally, insert the stirrer in the second hole of the lid with the loop encircling the bulb of the thermometer. Wait approximately 5 minutes before recording the temperature of the water to the nearest 0.1 oC.

2. Measure the mass of a weighing boat, add approximately 30 g of a known sample of metal, and obtain the mass of the metal by difference. (The metals used should be large pieces of metal; metal shot and foil don’t work very well.) Transfer the sample into an appropriately sized test tube, ensuring that no sample is lost.

3. It is recommended that a trial run be performed in conjunction with your partner before data is actually recorded for steps 4 and 5. You actions need to be swift and accurate to ensure quality data (and results!), but safety cannot be compromised. Take the time to familiarize yourself the procedure and determine the best way to execute it.

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4. To a 250-mL beaker, add approximately 150 mL of water and bring to a rolling (but not violent!) boil. Once the water is boiling, immerse the test tube containing the metal into the water using a clamp to support the test tube. After ~10 minutes in the boiling water, the sample inside the test tube and the boiling water should be at the same temperature. Record the temperature of the water to the nearest 0.1 oC. Transfer the thermometer to the calorimeter, and allow it to come to the same temperature that was recorded earlier.

5. Quickly transfer the hot metal in the test tube to the calorimeter containing the water. Take care to not splash any water out of the calorimeter! Immediately replace the lid, thermometer and stir wire, and record the temperature at time zero. Gently stir the water with the wire and record the temperature in 15 second

intervals until the temperature decreases for at least five measurements. This may take a few minutes. It is extremely important that the water be constantly stirred throughout the experiment. Finally, record when the temperature is at a maximum, which will be immediately before the temperature begins to descend.

6. Repeats steps 1-5, to determine if you obtain the same value for the specific heat of the known metal. Return your metal sample to the waste container provided.

7. Using your data, compute the specific heat capacity, Cs, of the known sample. Also report the percent error from the value listed in Table 1. The specific of water, Cs,water , is 4.18 J/gºC, and the heat capacity of the calorimeter, Ccal, is 8.36 J/oC.

-qmetal = qwater + qcal

-(Cs,metal × mmetal × Tmetal) = (Cs,water × mwater × Twater) + (Ccal × Tcal)

T = Tf - Ti

B. Measure the Cs of an Unknown Sample.

8. Use the same procedure for determining the specific heat of a known sample to determine the specific heat of an unknown sample. Be sure to indicate which unknown you have.

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LAB REPORT

Specific Heat Capacity

Name ________________________________ Date _________ Report Grade ______

Partner ________________________________ Section _________

A. Specific Heat of a Known Metal.

1. Mass of the coffee-cups ____________

Mass of the cups + water ____________

Mass of the water ____________

Temperature of cool water ____________

2. Identity of metal sample ____________

Mass of metal sample ____________

Temperature of boiling water ____________

3. Time vs. temperature data – first run:

Time(sec)

Temperature(°C)

Time(sec)

Temperature(°C)

Time(sec)

Temperature(°C)

0 105 210

15 120 225

30 135 240

45 150 255

60 165 270

75 180 285

90 195 300

Maximum System Temperature (Tf) ____________

4. Calculate the specific heat of the known and percent error . Show work!

Cs = __________ J/g∙oC % error = ____________

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5. Mass of the coffee-cups ____________

Mass of the cups + water ____________

Mass of the water ____________

Temperature of cool water ____________

6. Identity of metal sample ____________

Mass of metal sample ____________

Temperature of boiling water ____________

7. Time vs. temperature data – second run:

Time(sec)

Temperature(°C)

Time(sec)

Temperature(°C)

Time(sec)

Temperature(°C)

0 105 210

15 120 225

30 135 240

45 150 255

60 165 270

75 180 285

90 195 300

Maximum System Temperature (Tf) ____________

8. Calculate the specific heat of the known and percent error . Show work!

Cs = __________ J/g∙oC % error = ____________

9. Average specific heat of the known and percent error .

Cs = __________ J/g∙oC % error = ____________

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B. Specific Heat of an Unknown Metal.

1. Unknown Number ____________

Mass of the coffee-cups ____________

Mass of the cups + water ____________

Mass of the cool water ____________

Temperature of water ____________

2. Mass of unknown ____________

Temperature of boiling water ____________

3. Time vs. temperature data:

Time(sec)

Temperature(°C)

Time(sec)

Temperature(°C)

Time(sec)

Temperature(°C)

0 105 210

15 120 225

30 135 240

45 150 255

60 165 270

75 180 285

90 195 300

Maximum System Temperature (Tf) ____________

4. Calculate the specific heat of the unknown and identify it. Show work!

Cs = __________ J/g∙oC Identity = ____________

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