This document summarizes a lab experiment to calculate the heat capacity of solids. It describes measuring the heat capacity of a calorimeter vessel filled with water, then adding solids like graphite and copper pre-heated to 200°C to calculate their specific heat capacities. The experimental procedure, data analysis and calculations are shown. The specific heat capacities calculated for graphite, copper and porcelain are compared to known values, with relative errors of 8.9%, 14.2% and 26% respectively, showing reliable results except for graphite. Sources of error like heat loss during measurements are also discussed.

1. PHYSICS ENGINEERING DEPARTMENT
FIZ341E - Statistical Physics and Thermodynamics
Laboratory
Name of Exp. : Heat Capacity of Solids
Date of Exp. : 10.12.2014
BARIŞ ÇAKIR
090100235

2. Hocam ben Sercan hocanın odasına bırakmıştım
raporumu ama tedbir olsun diye buraya da atıyorum.
Introduction
Heat Capacity: is required heat energy to increase temperature of a certain
amount of matter for 1 o
C. It can be calculated with the following formula;
=
In this formula, c is specific heat and M is mass also heat capacity is different
for every matter. The theory of experiment comes from the energy conservation
law of closed systems, all formulas and calculations, extracted from this
principle;
=
According this principle, aim of the experiment is adding some materials to the
calorimeter vessel, for producing heat transfer, measuring temperature change
and calculating heat capacity values for each matter; however, calorimeter is
also have a heat capacity and it can not be neglected. So at the beginning of the
experiment, heat capacity of vessel is calculated with adding cold and hot water
in it (just heat transfer between hot and cold water).
= − − ( − )
Heat capacity can be extracted from formula above;
=
− − ( − )
−
We also know released heat equals, multiplication of specific heat, mass, and
temperature change so,
=
( − )
Released heat equals, absorbed heat so if we plug in this definition to equation
above;
=
−
( − )

3. Specific heat can extracted. Finally to reach specific heat capacity expression we
need to add heat capacity of vessel, and our equation will be like,
=
+ −
( − )
Specific heat is different for each matter so, by final equation, type of material
which is used in experiment can be founded.
Experimental Procedure
Tools and devices: Calorimeter vessel, kettle, balance, water,
thermometer, graphite, copper and porcelain.
First, experiment started with calculating heat capacity of the calorimeter
vessel, cold water added to vessel mass of vessel and water is calculated. We
add hot water on cold water and equilibrium temperature is measured and heat
capacity of vessel calculated with the equation which is given at information
part.
Second, specific heat of some solids calculated. We already got hot and
cold water mix at equilibrium temperature from first part, with putting solids
into this water, we calculated their mass and equilibrium temperature of solid
and water. Notice that all solids are heated to 200 o
C. With measured mass of
solid and equilibrium temperature of solid water mix specific heat of solid can
be calculated. This step is repeated for copper, porcelain and graphite.
Finally, absolute errors calculated with comparing experimental and real
values.
Data Analysis
The weight of vessel without water is,
= 865.45
when we add the cold water,
= + = 1002.75
cold water’s mass can be extracted as m,
= 137.30 .
After 10 minutes equilibrium temperature of cold water measured as,

4. = 19
we heated another water with kettle,
= 97
and we pour it in the vessel, equilibrium temperature of hot and cold water mix
is measured as,
= 44
and also hot water’s mass measured by, extracting total mass of vessel with and
without hot water,
= 1094.10
= − = 91.35 .
Heat capacity of the vessel can be calculated with all of these measurements and
heat capacity formula which is given at introduction part;
=
1 ∗ 91.35 ∗ (97 − 44) − 137.30 ∗ 1 ∗ (44 − 19)
44 − 19
= 56.362 /
After finding heat capacity absolute error (∆C) and relative error (∆C/C) can be
calculated;
(∆ ) = 39,64
∆
=
39,64
56.362
= 0,70
For, specific heat of solids we prepared a table for measurements,
Porcelain Copper Graphite
Mwater(g) 1094.15 1093.25 1080.1
T2 44 44 42
T1 200 200 200
Tf 45 45 43.5
Msolid(g) 31.1 25.35 9.55
c (joule/gC) 0.23867 0.3304 1.0402
Table1. Measurements with solids

5. T2 - initial temperature of vessel
T1 - initial temperature of solid
Tf - equilibrium temperature
Msolid - mass of solid
Mwater - initial mass of water
Real specific heat values of solids are,
= 0.82 /
= 0.385 /
= 0.26 /
Conclusion
We calculated all of specific heat capacities of used metals in the
experiment, by calculating the temperature difference and calculating heat
difference. According to error percentages we can discuss our result’s reliability.
% = %26
% = %14.2
% = %8.9
Except graphite we get reliable results, the issue at graphite can be explained
with number of graphite used in experiment, we put many graphite and
calculated it as one big graphite this can cause some extra error rate. Also we did
not calculate heat loss between air and solid, when we take solid out of the oven
it starts to translate heat with air, we know temperature difference between air
and solid is too high when we take solid from the oven, so there is a huge loss of
heat in this phase. For a perfect result relative error should equal to 1, our
relative error is 0.70 so we can say, we calculate heat capacity of calorimeter
vessel successfully.
Resources
 Thermodynamics LAB FÖY
 https://www.google.com.tr/search?q=heat+capacity+of+solids+experimen
t&source=lnms&tbm=isch&sa=X&ei=T2uIVNiPHYWBU-
Xlg5gO&ved=0CAgQ_AUoAQ&biw=826&bih=883
 http://www.engineeringtoolbox.com/specific-heat-solids-d_154.html

6. Answers
1- In this condition T1 is initial temperature and T2 will be the equilibrium
temperature, we can guess which elements used by calculating specific
heat, and we can calculate how much energy taken from T1 to T2 with
following formulas;
= ( 2 − 1)
=
We know T2,T1 and heat capacity values so we can calculate how much energy
taken from T1 to T2 with first formula and second formula will give which
material we used.
2- In this question we can make assumption on same formula;
= −
= ( − )
The subject is about cooling times, so we can say, relation between final
temperatures are,
>
So same heat given to both, they got same mass and initial temperature so,
>
This condition explains, toffee’s long cooling period.