Flames, Heat and Calories

FLAMES, HEAT AND CALORIES 8

This experiment was conducted with the idea of thermodynamics inmind. This is due to the fact the experiments were rife withprocesses that sought to find out relations between energy, work andthe transfer of heat. There were various aims and goals for thisobjective which were finding out the amount of heat gained or lost bymoving cold or hot water. This experiment also aimed at finding outthe amount of heat in metals. It is also essential to state that theexperiment aimed at finding out the variations in energy resultingfrom the chemical burning of a candle. Lastly, it is worth statingthat the experiment aimed at recording the physical changes ofmelting ice due to heat changes.

Thermodynamics is a term that can be defined through its two majoraspects which are heat and temperature (Rao, 2014). The differencebetween temperature and heat energy is that temperature is the amountof energy in a substance when heat is a form of energy exchangedbetween to substances due to variations in temperature. Research hasindicated that there are a number of ways of measuring energy.However, it is worth noting that whichever method one uses, energycan only be in the form of kinetic or potential energy. Whereas thereare various temperature units such as calories, joules and degreesCelsius, this experiment relied on calories as the unit of measuringthe heat (Rao, 2014).

This is an experiment that sought to find out the amount of heat offusion that was required to melt 1 gram of ice which was used as thesolid substance. Through the experiment, it was clear that there is adifference between the heat of fusion and the heat of vaporization.Heat of fusion is the amount of heat that is sufficient to melt onegram of a solid substance whereas heat of vaporization is amount ofheat sufficient to change a liquid into gaseous state (Rao, 2014). Itis useful to state that the experiment helped to identify both thephysical and the chemical changes of substance. A physical change islike the change in state such as liquid to gas whereas chemicalchange is like the changes in the candle after burning. The aspect ofthermodynamics is applicable I real life in instances such as whensausages are freezed in a refrigerator and they defrost when removedafter gained energy from the surrounding (Rao, 2014).

Data and results:

According to the findings of the experiment, it is evident thatthere was transfer of energy from the hot to the cold substances.This was the same case when hot water was mixed with cold water.Additionally, the experiment indicated that heat of burning a candlewas more than that of burning the marshmallow. Calculations indicatedthat the heat produced by the candle was 7 times more than thatproduced by the marshmallow. When ice is added to warm water, thewarm water loses energy than it gained. This was 10 times more thanwhat was gained. On the experiment of the unknown metal, it wasevident that the metal lost 26.21 degrees in temperature while thecold water gained 7.7 degrees in temperature.

Mass of cup

1.947g

Mass of cup and cold water

33.990 g

Mass of cold water

32.043g

Mass of cup and mixed hot and cold water

72.449g

Mass of hot water

29.613g

Table 1 above is from experiment 1.

Heat gain calculation:

Cold water sample

Hot water sample

Temperature before mixing

12.40 C

48.00 C

Temperature after mixing

23.78 C

Change in temperature

11.36 C

-2.92 C

Table 2 is from experiment 1.

Heat gain = (specific heat of water)(mass of cold)(temperaturechange)

= ((1)/(g*C))x (32.043) x (11.36 C) =364.0084 cal

Heat loss calculation:

Heat loss = (specific heat of water)(mass of hot water)(temperaturechange)

((1 )/(g*C))x (29.613g) x (2.92 C) = 86.470 cal

Ratio (Heat gain)(Heat loss):

(Heat gain)/(Heat loss) = 364.0084/86.470 = 4.21 cal

-Mass and temperature for water, candle, and marshmallow

Candle

Marshmallow

Mass of cup and water

107.668g

108.231g

Mass of empty cup

10.668g

10.668g

Mass of water

97g

97.563g

Water temperature before heating

19C

12.58C

Water temperature after heating

34.33C

14.67C

Temperature change

15.33C

2.09C

Mass of candle and metal plate or marshmallow and screen before burning

7.399g

60.407g

Mass of candle and metal plate or marshmallow residue and screen after burning

7.078g

60.222g

Mass of candle or marshmallow burned

0.321g

0.185g

Table 3. The tableaboverepresentsthedatathat wasobtainedfromexperiment part 2

Calories of heat from candle burning

(Heat gain) = (specific heat of water)(mass of coldwater)(temperature change)

x(97 g) x (15.33 C) = 1487.01 cal

Calories of heat produced per gram of candle wax burned.

1487.01 cal / 0.321g= 4632.43

Calories of heat from the burning marshmallow absorbed by the water

x(97.563) x (2.09 C) = 203.9 cal

Calculate calories of heat produced per gram of marshmallow burned.

(Heat loss) = 203.9 cal / 0.185 g = 1102.162 cal/g

Part 3

-Mass and temperature

Mass of cup and water

99.94 g

Mass of cup

1.778g

Mass of water

98.162g

Mass of cup and water and ice

108.152g

Mass of cup and water

99.94g

Mass of ice

8.212g

Temperature of water before adding ice

33C

Temperature after ice melted

25.15C

Temperature change of water sample

10.85C

Temperature after ice melted

25.15C

Initial temperature of ice

0.0C

Temperature change of melted ice sample

25.15C

Table 4. The table above represents the data that was obtained fromexperiment part 3.

Heat loss calculation:

Heat loss = (specific heat of water)(mass of hot water)(temperaturechange)

Heat loss =1 x(98.162g) x (10.85C) = 1065.0577cal

Heat gain:

(Heat gain) = (specific heat of water)(mass of coldwater)(temperature change)

=(1 cal/gC) (8.212) (25.15 C) = 206.53 cal

(Heat needed to melt 1 g of ice) = (1065.0577 cal) – (206.53 cal) =858.5277 cal

(Heat of fusion) = = = 104.546 cal/g

Part 4

– Heats of metal

Metal 1

Mass tested

Unknown

Mass of metal sample

70.152g

Mass of nested Styrofoam cups and cold water

42.224g

Mass of nested Styrofoam cups

3.786g

Mass of cold water sample

38.438g

Temperature of cold water before transfer

18.51 C

Maximum temperature of cold water after transfer

26.21C

Change temperature of cold sample

7.7C

Temperature of metal before transfer

88.65C

Maximum temperature of cold water after transfer

26.21 C

Change temperature of metal sample

62.44 C

Calculate specific heat

0.763J/g- C

Accepted value for specific heat

0.3846J/g- C

Table 5. The table above represents the data that was obtained fromthe experiment part4.

-Unknown:

Heat loss = (specific heat of water)(mass of cold water)(temperaturechange)

x(38.438g)x (7.7 C) = 1238.349 J

(Specific heat) =

(Specific heat of water)(Mass of cold water)(Temperaturechange)=(specific heat of metal)(mass of metal)(change intemperature)

(Specific heat) = x (38.438g) x (7.7 C) = (X) x (70.152g) x (7.7 C) = 2.2925J/g*C

Conclusion

This is an experiment that lived up to its expectations. Thecalculations of heat gains, losses and heats of substances were alfound. The chemical and the physical changes were also identifiedthrough the experiment. Change of state of the ice to liquid waterwas a physical change while the change of the marshmallow waschemical change. There were, however, a few errors that wereidentified in the experiment such as the loss of heat to thesurrounding and the poor thermometer calibration. It is worth notingthat the objective of the relationship between heat systems and thesurrounding was achieved.

Reference

Rao, Y. V. C. (2014).&nbspAn introduction to thermodynamics.Hyderabad: Universities Press.