Specific Heat and Latent Heat Lab

EXPERIMENT NUMBER: #13

EXPERIMENT NAME: Specific Heat and Latent Heat

EQUIPMENT: Metal cylinder, Triple beam balance, thermometer, Beaker, Hot plate, and a Styrofoam cup.

OBJECTIVE: The purpose of this experiment is to measure the specific heat capacity of an unknown metal samples and also to determine the latent heat of fusion of ice.

THEORY: My partner and I plan to measure the specific heat of a block of metal and the heat fusion of ice. We will do this by using the tools listed above and the equation listed below.

Q=C∆T=cm∆T

∆Q=mL

C_(p,s)=(M_(H_2 O) C_(p,H_2 O) (T_f-T_i ))/(M_s (〖100〗^o C-T_f ) )

L_fus=C_(H_2 O) (M_(H_2 O)/M_ice (T_i-T_F )-(T_f-0^o C) )

PROCEDURE: The first thing me and my partner did was put about 100 ml of water in a beaker and put it on the hot plate to bring the water to a boil. Then we measured the mass of our block with the triple beam balance scale. After that we placed the metal object in our boiling water. While that was in the water we measured the mass of our cup with the triple beam balance scale and put about 100 ml of water inside the cup. We then calculated the mass of the water in the cup and also measured the temperature of the water in that cup with a thermometer. We then added our mass that had been boiling, into the cup filled with tap water and found our T max. We did that step two more times to complete that experiment. After that we found the specific heat for all the trials in Cgs units and converted it to SI units using the conversion factor 1 Cal/g = 4186 J/kg. We then moved on to the second experiment, which was finding the latent heat of fusion. With a known mass of the water in the cup at about 60-66 degrees Celsius that we measured with the thermometer, we then took ice and placed it in our cup. We kept adding ice until the temperature reached between 8-10 degrees Celsius and then recorded it. We then measured the final mass of the cup we put the ice in to with the triple beam balance scale, to see how much ice we added to the total mass. We then did these steps two more times and then calculated the latent heat of fusion.

CALCULATIONS:

Specific Heat

168.6(1)(30-21)/(86.9(100-30))=.2494 cal/〖g 〗^o C

.2494*4186=1044.1947 J/(〖kg 〗^o C )

172.4(1)(29-20)/(86.9(100-29))=.2515 cal/〖g 〗^o C

.2515*4186=1052.6909 J/(〖kg 〗^o C )

203.2(1)(28-20)/(86.9(100-28))=.2598

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