Determining the Specific Heat Capacity of an unknown Metal

Angela Tan

Tasnim, Priya, Marie, Pedram, Nick, Ken, and Eric

May 18, 2011

Abstract

The purpose of this experiment is to determine the joules per kilograms (the specific heat capacity) of a metal object and to identify that metal based on its specific heat. By doing so, a zero degree unknown metal was put into room temperature water. The metal was left in the water until equilibrium. The GLX was used to measure the temperature of the water containing the metal. In the end results were gathered to determine the unknown metal based on its specific heat.

Introduction

To find the specific heat capacity of the unidentified metal: the mass of the metal was needed, mass of water, initial and final temperature of metal and water. The heat capacity of water was already given. The initial temperature of the metal was measured with the GLX after it was in the ice cold water and reached a temperature of zero degrees. The initial temperature of the water was measured at room temperature in a cup with the GLX. The final temperature of the metal and water was measured when the temperature stopped changing and has reached a stable temperature. Temperature changes depend on the thermal energy flowing into and out of an object. The change varies from the mass of the object, what it is made of, and the temperature of the environment. “The amount of thermal energy that a single gram of a specific material must absorb in order to change its temperature by one degree is the material’s specific heat capacity.” Once the results were accumulated and put into the equation, the specific heat capacity can be calculated.

Theory

The thermal energy that was lost or gained on the metal equals the thermal energy that was gained or lost by water. This equation explains the equilibrium of metal and water:

Qlost=Qgaine d

Qlost is the amount of thermal energy that was lost and Qgained is the amount gained onto another object due to the lost of the other object.

[1]

The change in thermal energy can also be written as:

∆Q=mc∆T

The ∆Q represents the change in thermal energy. It depends on the mass, m, in kilograms, the specific heat, c, in joules per kilogram celcium, and the change in temperature, ∆T. This is to measure the heat transfer amount.

“When an unknown metal object is put in the water in acalorimeter, the change in thermal energy of the object equals the change of thermal energy of the water.” The initial temperature of the unidentified metal and water will be different, but in the end it will be the same – called equalibrium temperature.

MobjectCobject∆Tobject = MwaterCwater∆Twater

This equation represents that the thermal energy between the two is equal. The symbol m is mass, c is the heat capacity, and ∆T is the change in temperature.

To figure out how far off the experimental result from the actual result, this formula was used:

%diff=|accepted−measuredaccepted |x 100

The ‘accepted’ is the actual heat capacity of the metal and the measured is what the results gave. It is multiplied by a hundred to figure the percent out of a hundred.

Apparatus and Procedure

The Fast-Response Temperature probe was plugged into the first temperature port and the GLX was turned on. A graph appeared on the screen that shows Temperature (oC) and Time (s). The mass of the unknown metal was determined by a scale and the PACO Basic Calorimetry Set. The given mass number that was in grams was converted into kilograms and then recorded onto the data table. A large beaker was filled with ice and water. The metal was wrapped with the braided physics string so it can be taken out of the ice cold water when it reached zero degrees. The initial temperature of the metal was taken. A Styrofoam cup was measured and recorded. The cup was after filled with tap water and measured with the scale and recorded onto the chart. All of the measurements were converted into kilograms. To find the mass of water that was in the cup, we subtracted the mass of the cup including water from the cup alone. The room temperature water was then put into the Calorimetry set along with the metal that was taken out of the water bath. The rod of the Fast-Response Temperature Probe recorded the temperature changes in the cup. When the temperature stopped changing, the final temperature of the metal and water was recorded.

[2]

[3]

[4]

Figure 1: This was the Xplorer GLX that was used in the experiment to measure the temperatures of the metal and water.

Data

Table 1: Results from the activity

Item ValueMass of Object 0.0558kg

Mass of cup 0.0015kgMass of cup plus water 0.1230kg

Mass of water 0.1215kgInitial Temperature of Metal 0oCInitial Temperature of Water 23.9oC

Final Temperature of Metal/Water 22.5oC

Analysis

To figure out the heat capacity of the unidentified metal equation [3] has to be used. The two heat transfer equation equal each other for the reason as to why metal and water meet at equilibrium.

MoCo∆To=MwCw∆Tw

MoCo(Tf-Ti)o=MwCw(Ti-Tf)w

Co=MwCw (Ti−Tf )wMo (Tf−Ti )o

Co=( .1215 ) (4186 )(23.9−22.5)

( .0558 )(22.5−0)

Co=712.03861.2555

Co=567.135 J/kg oC

Table 3: Specific Heat of the Metal

Item ValueSpecific heat of metal 567 J/kg oC

Table 4: Specific heats of common metals

Metal Specific Heat (J/kg oC) Metal Specific Heat (J/kg oC)Aluminum 901 Iron 449Brass 380 Lead 128Copper 386 Silver 234Gold 129 Zinc 387

According to table 4 the result is closest to iron.

To solve the percent error, equation [4] has to be used to solve the percent difference.

%diff= |accepted−measuredaccepted |x100

%diff= |567−449449 |x100

%diff= |118449|x100

%diff=26.3

The ‘accepted’ and ‘measured’ had to be reversed otherwise the answer will be negative. This equation will work perfectly if the measured was lower than the accepted. In this case the measured was higher than the accepted.

Conclusion and Discussion

Thermal equilibrium only occurs when there is two different initial temperatures so heat transfer can happen. When there is no heat transfer there is no change in temperature and no change of temperature, there is no thermal equilibrium. For example if you were cold and entered a warm room, you will start to feel warm over time. That is heat transfer. If you were warm and entered a warm room, you don’t feel a change of temperature because there is no change of temperature. The metal that was identified was iron. It may not have the exact specific heat, but it was semi close to the actual results. The experimental results could have been better if there were more trials of the

activity, which gives a better chance of a greater result. Factors that could have contaminated the experiment were: not waiting long enough for the water and metal to reach equilibrium; the metal that came right out of the water bath got warmer; or the water could have been contaminated. Those factors could have changed a fraction of the experiment which could really change the whole entire heat transfer equation. In result of the experiment it we could have assumed or hypothesized the wrong metal based on in accurate information. The percent difference was rather small and for that reason the experiment went adequate. If it was more accurate