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Experiment #4: Specific Heat of a Metal
Data and results
Data table:
Complete and submit the highlighted sections to your instructor before the end of the class session.
Aluminum (Al)
Your Metal Choice: _______
Mass of metal (g)
Volume of water (mL)
Mass of water (g)
Initial temperature of water (°C)
Initial temperature of metal (°C)
Final temperature of water + metal (°C)
Results table
Aluminum (Al)
Your Metal Choice
Change in temperature (ΔT) of water
Heat absorbed by water
ΔT of metal
Specific heat of metal
Show all calculations for the
Specific heat of Aluminum
below:
Jachimma Onwuamaegbu
09/24/2022
Bronze
6.7870g
19.979
100.0 mL
100.0 mL
98.8g
98.9 g
25.00 C
25.00 C
200 C
200 C
27.38 C
26.75 C
2.38 C
1.75 C
67.52 J
146.29 J
172.62 C
172.62
0.06 J/g C
0.42 J/g C
Heat absorbed (Al): Q= m*c*change in T
Q= 6.7870g* 4.184J/g degrees C* 2.38 C= 67.52
Heat absorbed (Bronze): Q= 19.979g * 4.184J/ g C *1.75 C
Q = 146.2862 J
Q= 146.29
Specific heat (Al)= C= Q/ (change in T * mass)
C= -67.52 J/ (-172.62 C * 6.7870g) = 0.05763 J/g C
= 0.06 J/g C
Specific heat (Bronze)= C= Q/ (change in T * mass)
C= -146.29 J/ (-172.62 C * 19.979g) = 0.4241 J/g C
= 0.42 J/g C
Postlab
Questions
1.
Differences in equipment used for obtaining measurements in the lab can introduce experimental errors
into calculated values. To minimize the effects of these experimental errors on our calculated values
we conduct multiple trials of an experiment and obtain an average of the values.
Calculate the percent
error in the specific heat value that you determined experimentally.
The accepted specific heat value
for Aluminum is 0.900 J/g°C.
%
𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒
=
�
𝑦𝑦𝑒𝑒𝑦𝑦𝑒𝑒
𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑒𝑒𝑒𝑒 − 𝑎𝑎𝑎𝑎𝑎𝑎𝑒𝑒𝑎𝑎𝑎𝑎𝑒𝑒𝑎𝑎
𝑣𝑣𝑎𝑎𝑣𝑣𝑦𝑦𝑒𝑒
𝑎𝑎𝑎𝑎𝑎𝑎𝑒𝑒𝑎𝑎𝑎𝑎𝑒𝑒𝑎𝑎
𝑣𝑣𝑎𝑎𝑣𝑣𝑦𝑦𝑒𝑒
� 𝑥𝑥
100%
2.
When conducting this experiment, you are usually instructed to transfer the metal quickly from the
oven or hot water bath to the calorimeter. Why would you need to do this?
_______________________________________________________________________________
_______________________________________________________________________________
________________________________________________________________________
__
_____
3.
A metal object with a mass of 1
45
.25 g is heated to 97.5 °C and then transferred to a calorimeter
containing 84.58 g of water at 2
1
.5 °C. If the water and metal reach a final (equilibrium) temperature
of 3
1
.
8
°C, what is the specific heat of the metal object? Show all work for full credit.
You are usually instructed to transfer the metal quickly from the oven or hot water bath
to the calorimeter because if you take a long time, the heat can be going out in the air.
Basically losing the heat.
% error = (0.06 - 0.900/ 0.900) * 100= -93.33%
Heat absorbed: Q= m*c*change in T
145.25g * 4.184J/ g C * 10.3 C = 6,259.58 J
C= Q/ (change in T * mass)
C= 6,259.58 J / (65.7 C * 145.25g) = 0.6559 J/g C
Specific heat of the object is 0. 7 J/g C
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