Science

Purpose: To find the specific heat capacity of a substance

Principle:  According to the principle of thermal energy exchange, when thermal energy is transferred from a warmer object to a colder object, the amount of thermal energy released by the warmer object is equal to the amount of thermal energy absorbed by the colder object.


Q released =Qabsorbed


The quantity of heat required to heat up a substance through T is Q = mc∆T

m1 c1 ∆T1 = m2 c2 ∆T2  ( or )  c1=  (m2 c2 ∆T2 ) / m1 ∆T1

Procedure:

Measure the mass of water

Measure the initial temperatures of water and cube

Drop the cube in water and note the final steady temperature of the system

Measure the mass of the cube

Repeat the experiment with different cubes

Calculate the temperature changes that happen to the water and the cube

Calculate the specific heat capacity of each cube

Observations:

Mass of water m2= 224g = 0.224 kg

Specific heat capacity of water c2 = 4200 J / kgC

 Initial Temp of water ( t2 ) CInitial Temp of cube ( t1 )  0CFinal Temp of the mixture ( T ) CMass of cube ( m1 ) kg
Cube 122.5C    -6C18C0.205 kg
Cube 220.5C    -6C20C0.066 kg
Cube 321C    -6C20C0.068 kg

Calculations:

 ∆T2 = t2 – T ( C ) ( Water )∆T1 = T – t1 ( C ) ( Cube )c1 = (m2 c2 ∆T2 ) / m1 ∆T1 J / kg C
Cube 122.5 – 18 = 4.5C    18 – (–6) = 24C    (0.224)(4200)(4.5)/(0.205)(24) = 861.6 J/kg℃
Cube 220.5 – 20 = 0.5C    20 – (–6) = 26C    (0.224)(4200)(0.5)/(0.066)(26) = 274.1 J/kg℃
Cube 321 – 20 = 1C    20 – (–6) = 26C    (0.224)(4200)(1)/(0.068)(26) = 532 J/kg℃

The calculated value of specific heat capacity of cube 1 = 861.6 J/kg℃

The calculated value of specific heat capacity of cube 2 = 274.1 J/kg℃

The calculated value of specific heat capacity of cube 3 = 532 J/kg℃

Actual Specific heat capacity of cube 1 =  880 J/Kg℃ (Based on Researched heat capacity for Aluminum)                             

Actual Specific heat capacity of cube 2 = 236 J/Kg℃ (Based on Researched heat capacity for Silver)

Actual Specific heat capacity of cube 3 = 521 J/Kg℃ (Based on Researched heat capacity for Titanium)

% error for cube 1 = (Observed – Actual)/Actual * 100% = (861.6-880)/880 *100% = 2.09 %

% error for cube 2 = (Observed – Actual)/Actual * 100% = (274.1-236)/236 *100% = 16.14 %

% error for cube 3 = (Observed – Actual)/Actual * 100% = (532-521)/521 *100% = 2.11 %


Conclusion:

In conclusion, the purpose of this lab was to determine the specific heat capacity of a substance by using the principle of thermal energy exchange and the equation Q = mc∆T. By measuring the temperature changes and masses of the substance and water in three separate experiments, we were able to calculate the specific heat capacity of the substance using the equation m1c1∆T1 = m2c2∆T2. The accuracy of our results may have been affected by various sources of error, such as heat loss to the surroundings and the assumption of a perfectly insulated system. Nonetheless, this lab provided a valuable opportunity to apply fundamental principles of thermodynamics and to gain practical experience with experimental techniques for measuring thermal properties.

author avatar
William Anderson (Schoolworkhelper Editorial Team)
William completed his Bachelor of Science and Master of Arts in 2013. He current serves as a lecturer, tutor and freelance writer. In his spare time, he enjoys reading, walking his dog and parasailing. Article last reviewed: 2022 | St. Rosemary Institution © 2010-2024 | Creative Commons 4.0

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