Vaughn College Physicsli Ka Name Date Specific Heat Specific Heat describes how

Question

Vaughn College Physicsli Ka Name Date Specific Heat Specific Heat describes how "sensitive" a material is to heat energy. For example, the specific heat (usually represented by the letter "c") of water is 1.00 cal / gC. This means if 1.00 calories of heat energy is added to 1 gram of water, its temperature will increase by 1 o C, and conversely, if 1.00 calories is removed the temperature will drop by 1C. In this lab you will find the specific heat of aluminum andsteet. PART Finding the specific heat of aluminum. Start by measuring the mass of the empty sample holding container (the cylinder with the 3 cm orrer diameter) Mass of cylinder is 3 Next fill this cylinder approximately ½ to2/3 of the way to the top with aluminum pellets. Mass of cylinder with pellts OK.E Calculate the mass of the aluminum pellets Mass of pellets (mp) Next prepare the calorimeter Mass of inner calorimeter cup m)- Fill the cup approximately ½-2/3 of the way to the top with water. Measure the mass of the cup with 53 the water in it Mass of cup with water-2D.Y Use the above two mass measurements to calculate the mass of the water. 86 Mass of water in calorimeter cup ( m w ) = Assemble the calorimeter. Be sure to place the lid on top, and insert the Vernier temperature probe thru the small opening in the lid. Be sure the probe is in contact with water. Log onto "logger po"; the temperature reading data should shortly appear on the computer screen. Record the initial temperature of the calorimeter water

Explanation / Answer

In the beginning,

For aluminum pellets,

Mass = 85.2 g

Temperature = 85 C

Specific heat = cp

For calorimeter cup,

Mass = 53.9 g

Temperature = 22.4 C

Specific heat = 0.215 cal/g C

For water in calorimeter,

Mass = 186.5 g

Temperature = 22.4 C

Specific heat = 1.00 cal/g C

When mixed,

Temperature = 27 C

Hence

Using energy conservation,

Heat lost by aluminum = heat gained by (calorimeter cup + water)

-mpcp(Tf – Tp) = mccc(Tf – Tc) + mwcw(Tf – Tw)

As Tc = Tw = Ti so

-mpcp(Tf – Tp) = mccc(Tf – Ti) + mwcw(Tf – Ti)

Hence equation for cp is,

cp = {mccc(Tf – Ti) + mwcw(Tf – Ti)}/{-mp(Tf – Tp)}

Experimental value is,

cp = {53.9*0.215*(27 – 22.4) + 186.5*1.00*(27 – 22.4)}/{-85.2*(27 – 85)}

cp = 0.1844 cal/g C

Standard value is,

cp = 0.215 cal/g C

Percent difference = {(0.215-0.1844)/(0.215)}*100 = 14.2326.

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