Question

Experiment 23 Determination of Specific Heats

In determining the specific heats, what is the:

A) Equilibrium Temperature ?

B) Specific Heat Capacity of Copper?

C) Specific Heat Capacity of Granite?

D) the percent error?

(Solve for A,B,C,D)

Information:

Initial Temperature :100C

Mass of Copper :165 g  

Mass of Granite :165 g

Initial Temperature of Water :38C  

Mass of Water :3700g

Final Temperature for Granite : 38.52 C and Final Temperature for Copper : 38.25 C

(Please show how you solve each portion of this problem )

Answer 1

Initial Temperature, T1 :100C

Mass of Copper, Mc :165 g  

Mass of Granite, Mg :165 g

Initial Temperature of Water, T2 :38C  

Mass of Water, Mw :3700g

Final Temperature for Granite, T : 38.52 C and

Final Temperature for Copper, T : 38.25 C

A)

The data tells us that the final temperature is 38.52 degree for granite and 38.25 degree for copper.

Equilibrium temperature, To = 38.52 degree Celsius for granite and Equilibrium temperature, To = 38.25 degree Celsius for copper.

B)

For copper:

Heat lost by copper = heat gained by water

Mc Cc (T1 – T) = Mw Cw(T-T2)

0.165 Cc (100-38.25) = 3.7 x 4200x (38.25-38)

Specific heat of copper, Cc = 381.3 J/kgK

C)

For granite:

Heat lost by granite = heat gained by water

Mg Cg (T1 – T) = Mw Cw(T-T2)

0.165 Cg (100-38.52) = 3.7 x 4200x (38.52-38)

Specific heat of graphite, Cg = 796.6 J/kgK

D)

Actual specific heat of copper, Cc1 = 385 J/kgK

Percentage error in the measurement of copper = (385-381.3) x 100/385 = 0.96 %

Actual specific heat of graphite = 706.9 L/kgK

Percentage error in the measurement of specific heat of granite =

                                                                            =(796.6-706.9) x 100/706.9

                                                                            = 12.7 %