Lab Report- Heat Capacity

LabReport- Heat Capacity

Conservationof energy

Energycan be defined as the ability or capacity for doing work. Theprinciple of conservation of energy states that energy cannot becreated nor destroyed but can be transformed from one state toanother. Energy may exist in various forms and may be transformedfrom one type to another. The law of conservation of energy can alsobe rephrased as the total energy of an isolated system that remainsconstant.

Purposeof the Lab

Laboratoriesaid in providing a solid foundation for the theory taught duringlectures. It helps the students to interact with experimentalapparatus and equipment as well as methods of data analysis. Also thelab helps to introduce methods of dealing with uncertaintiesencountered during experiments. It also aids in distinction betweensystematic and random errors thus the importance of laboratoriescannot be under looked.


Partone:Heatloss and Heat gain

Twotemperature sensors were connected to the micro LAB and those sensorsnamed Temp A and B.

Adry empty Styrofoam cup was weighed and about 30ml of cold tap waterput in the cup. A little ice was added to the water so that when itmelted and the temperature would lower. The cup was reweighed todetermine the total mass of the cold water and the cup.

About30ml of hot water were added to another cup and one of thetemperature sensors put in each cup (A containing hot water and Bcontaining hot water) and the program started. Results were obtainedand recorded.

Parttwo:Heatassociated with chemical change

Aring stand was set up with a clay triangle on an iron ring to hold ametal cup above a candle. Carbon was cleared off the metal cup andwas weighed.100 ml of cold water was put in the cup then re-weighedto find the mass of the water in the cup. Temperature A sensor wasplaced in the water in the cup and the program modified so that thetemperature A data was taken and recorded. For the experimentinvolving peanut butter the above procedure was followed using peanutinstead of candle.

Partthree:Heatassociated with a physical change

AStyrofoam cup was weighed and then 100ml of warm water was added andweighed again to determine the exact mass of water. Temperaturesensor A was placed in the warm water. The temperature measurementprogram was run and then the ice dumped into the Styrofoam cup withthe warm water. Temperature of the water/ice sample was monitored andstirred until the ice had melted. The cup was taken to the balanceroom and weighed to determine the weight of the ice added.

Partfour:Specificheats of metals

Ametal sample (iron) labeled as A was obtained and the metal sampleweighed before it was attached to the middle of a tongue depressorwith a piece of string. A paper towel was used to dry two cups whereone was put inside the other and both weighed. A 100ml beaker wasfilled to about three- fourths with hot water and heating was startedwith a Burnsen burner. While water was still heating the program wasmodified to measure time, temp A and temp B. Upon the watertemperature reaching 85°C the Burnsen burner and both sensorsrecalibrated using the then heated water. The data was saved and theprocedure repeated with a different metal sample in that casealuminum.

Heatloss and heat gain

Thereare three modes of heat transfer namely convection,conduction and radiation.Heat transfer is the exchange of thermal energy which takes placefrom a region of high temperature to that of lower temperature. Astate of thermal equilibrium is attained when the bodies and thesurroundings attain the same temperature. The experiment where coldand hot water were mixed verified this concept since there wastransfer of heat from the system which had higher temperature, thatis hot water at 74.667°C, to that which had lower temperature thatis cold water at 14.373°C.

Uponmixing the hot and cold water, the mixture obtained a temperature of36.10°C. From the data obtained it was possible to calculate theamount of heat gained by the cold water as follows


PARTONE:Heatloss and Heat gain


WhereQ= Amount of heat

M=Mass (of water)

C=Specific heat capacity (of water)

ΔT=Temparature difference

Thespecific heat capacity of water is 1.000cal/g/°C. And from theexperiment it was found that the temperature difference was 21.727 asfollows, final temperature- initial temperature.



Onthe other hand the amount of heat lost by the hot water was found tobe

=20588x 1.000 x (-38.567). The negative sign indicates that there was lossof heat.


Theratio of heat gain/heat loss was calculated as follows


Fromthe experiment of mixing hot and cold water it was found that aportion of heat was inevitably lost to the surrounding. This is dueto the existence of a temperature difference between the two systems.

PARTTWO:Heatassociated with chemical change

Heatfrom burning candle= heat absorbed by water

FromQ=MCΔT it was found that heat absorbed by water= 100.06×1.00×15.371=1538.022Cal.

Thecalories of heat produced per gram of candle wax burned =1538.022/0.251= 6127.577

Caloriesof heat from burning peanut absorbed by water= 105.61×1.00×3.674=388.011Cal.

Caloriesof heat produced per gram of peanut burned=388.011/0.237= 1634.890

Fromthe experiment it can be deduced that the candle wax is a bettersource of heat than peanut butter and this can be seen from thecalories values of heat from burning candle and peanut respectivelywhere candle wax had a higher value.

PARTTHREE:Heatassociated with a physical change

Heatloss calculation= specific heat of water x mass of warm water xtemperature change

1.000×99.805 x 8.828 = 881.079Cal.


Theamount of heat required to warm the melted ice from 0 to the finaltemperature is

Heatgained by melt ice= specific heat of water x mass of ice xtemperature change

1.00×8.695 x28.003= 243.49Cal

Thedifference between the heat lost by the warm water sample and theheat gained by the melted ice is the amount of heat that went intothe melting ice.

Heatgained to melt ice= heat lost from warm water – heat gained bymelted ice

881.079– 243.49 = 637.589Cal.

Heatof fusion was calculated as follows heatgained to melt ice/mass of ice.Thus 637.589/8.695=73.34 Cal/g. The heat of fusion data obtained wasa close match to the accepted value of 79.7 Cal/g.

PARTFOUR: Specificheat of metals

Metal1 (Iron)

Heatgain calculation= mass of cold water x specific heat of water xtemperature difference 38.546 x 1.000 x 1.593 = 61.40Cal which isequivalent to (61.40 x 4.184) = 256.9Joules

Heatlost from the metal = heat gained by cold water

15.231x specific heat x -68.844 = 256.9J thus specific heat = 256.9/(15.231x – 68.844) = – 0.245

Thenegative sign is dropped as it shows that heat is lost. Thus theobtained value was 0.245J/g/°C

Metal2 (Aluminium)

Heatgain calculation = 38.253 x 1.000 x 0.685 = 26.20Cal which isequivalent to (26.20 x 4.184) = 109.63Joules

Heatlost from the metal = heat gained by cold water

16.300x specific heat x -61.630 = 109.63J thus specific heat =109.63/(16.300 x -61.630) = -0.109J/g/°C


Theobjectives of this experiment were to understand the concepts ofcalories and specific heat. It was subdivided into performing heatgain and heat loss calculations comparing the energy changesinvolved in chemical and physical changes and also to measurespecific heats of metals.

Theexperiment conducted was a huge success since all the objectives wereachieved. Also the measurements obtained in the laboratory were closeto the value contained in literature. The possible source of errorwas some heat loss to the environment. Also the specific heat of Ironand Aluminum had a slight disparity with the literature value andthis could be attributed to presence of impurities in the samples.

Arecommendation is that measurements should be made with extra cautionand the reagents to be handled with care so as to avoid contaminationduring handling.