PASCO Specialty & Mfg. ME-8569 User Manual download pdf (Page 16)

scientific

012-04774A

3. Watch the temperature of the cold water rise, stirring the

water gently. After several minutes, the temperature will

peak out (above room temperature) and began to fall.

Record this equilibrium (peak) temperature to the nearest

tenth of a degree in Table 1.

4. After the equilibrium temperature has been reached,

remove the object from the water and weigh the water with

the cup. Subtract the mass of the cup from the mass of the

water with the cup and record the mass of the water in

Table 4.1.

5. Calculate the change in temperature for the object:

∆T = HOT TEMP - EQUILIBRIUM TEMP.

Record this in Table 4.2.

6. Calculate the change in temperature for the water in the

styrofoam cup:

∆T = EQUILIBRIUM TEMP - COLD TEMP.

Record this in Table 4.2.

7. When the object is cooled in the water, the heat lost by

the object is equal to the heat gained by the water because

energy is conserved (neglecting any losses to the environ-

ment):

∆Q

OBJECT

= ∆Q

WATER

(mc∆T)

OBJECT

= (mc∆T)

WATER

Solving for the specific heat of the object gives:

c =

where c

WATER

is 1 cal/g °C. Calculate the specific heat for

the object and record the results in Table 4.2.

8. Look up the accepted value for the specific heat for the

material used and record in Table 4.2. Calculate the percent

difference between the experimental value and the accepted

value.

9. Repeat this procedure for the other objects.

QUESTIONS

1. Do the two aluminum objects have the same specific

heat?

2. Do the two brass objects have the same specific heat?

3. How does starting the cold water below room temperature

minimize the effect of the losses to the environment?

OBJECT OBJECT ∆T WATER ∆T

SPECIFIC

HEAT

ACCEPTED

VALUE

DIFF.

AL. CYL.

AL. BLOCK

BRASS CYL.

BRASS BLOCK

Table 4.2: Calculations

Figure 2: Equilibrium Temperature

WATER

(∆T)

WATER

OBJECT

(∆T)

OBJECT

Thermometer

Submerge mass

in water

1 2 ... 11 12 13 14 15 16 17 18

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