Gözde ÇEVİK
PHYSICS LAB: GÖZDE ÇEVİK
SPECIFIC HEAT CAPACITY OF WATER
INTRODUCTION
Research Question: How many joules of heat are required to raise the temperature
of 0.5 liters water by 1℃?
Aim: To determine the specific heat capacity of water.
Background Information:
Specific heat capacity is the amount of heat needed to raise a pure substance’s
temperature by 1℃. The following equation is used to find out the specific heat
capacity of any substance including water:
∆𝑄 = 𝑚 𝑥 𝑐 𝑥 ∆𝑡
Where
 ∆𝑄 means the amount of heat applied,
 m means the mass of the substance tested
 c means the specific heat capacity of the substance tested
 ∆𝑡 means temperature rise.
Gözde ÇEVİK
In this experiment, certain amounts of water placed in a kettle are heated for
specified time intervals before and after their temperatures are measured. The power of
the kettle is learnt before the experiment to measure the amount of heat applied (∆𝑄)
and the power of it is 1000 watts.
Hypothesis: Since the specific heat capacity of a substance is the same for all
conditions, the 4 different testings will give the same/close value of specific heat
capacity of water under the same conditions.
Independent Variables:
1. The amount of water used (0.5 liters)
2. The initial temperature of water used
3. The power of the boiler used (1000 Watts)
Dependent Variables:
1. The temperature rises of water for each case
2. The final temperatures of water for each case
3. The amount of heat applied.
Controlled Variables:
1. Time
2. Room temperature
How the variables are controlled: The whole experiment is done in a place that
was in room temperature and had the same air conditions throughout the experiment.
Gözde ÇEVİK
DESIGN
Apparatus:
1. Tap water
2. Graduated kettle
3. Centigrade thermometer
Method:
1. The power of kettle is noted down before the experiment.
2. Half liter of water is poured from the tap to the kettle for the first measurement.
3. The temperature of the water in the kettle is measured by using a thermometer.
4. The initial temperature is recorded
5. The kettle is turned on and the water is heated for 30 seconds.
6. After 30 seconds, the kettle is turned off and immediately the final temperature
of water is recorded.
7. The steps 2-5 are repeated for the second, third and fourth measurements but
the time for heating is changed to 60 , 90 and 120 seconds respectively.
8. The values are recorded.
Gözde ÇEVİK
DATA AND PROCESSING
Raw Data
Table 1: The initial and final temperatures recorded for the four trials.
INITIAL TEMPERATURE (℃ ±
FINAL TEMPERATURE
0.5)
(℃ ± 0.5)
1st trial
27 ± 0.5℃
37 ± 0.5℃
2nd trial
23 ± 0.5℃
47 ± 0.5℃
3rd trial
17 ± 0.5℃
53 ± 0.5℃
4th trial
13 ± 0.5℃
63 ± 0.5℃
TESTINGS
Processed Data
Table 2: The temperature rise of the four trials
Initial Temperature
Final
Temperature
(℃ ± 0.5)
Temperature (℃ ± 0.5)
Rise (℃ ± 0.5)
1st trial
27 ± 0.5℃
37 ± 0.5℃
10 ± 0.5℃
2nd trial
23 ± 0.5℃
47 ± 0.5℃
24 ± 0.5℃
3rd trial
17 ± 0.5℃
53 ± 0.5℃
36 ± 0.5℃
4th trial
13 ± 0.5℃
63 ± 0.5℃
50 ± 0.5℃
TESTINGS
Finding out the specific heat capacity of water:
In order to find out the specific heat capacity of water, the energy applied to all
trial samples must be measured first.
Gözde ÇEVİK
 𝑒𝑛𝑒𝑟𝑔𝑦 = 𝑝𝑜𝑤𝑒𝑟 𝑥 𝑡𝑖𝑚𝑒
1st Trial:
1000 W x 30 sec. = 30000 J is ∆𝑄 1
∆𝑄 = 𝑚 𝑥 𝑐 𝑥 ∆𝑡
c=
2nd Trial:
30000
0.5 𝑥 10
𝑚 𝑥 ∆𝑡
1000 W x 60 sec. = 60000 J is ∆𝑄 2
c=
60000
0.5 𝑥 24
so c =
∆𝑄
𝑚 𝑥 ∆𝑡
= 5000 joule/ gram ℃
1000 W x 90 sec. = 90000 J is ∆𝑄 3
∆𝑄 = 𝑚 𝑥 𝑐 𝑥 ∆𝑡
c=
4th Trial:
∆𝑄
= 6000 joule/ gram ℃
∆𝑄 = 𝑚 𝑥 𝑐 𝑥 ∆𝑡
3rd Trial:
so c =
90000
0.5 𝑥 36
so c =
∆𝑄
𝑚 𝑥 ∆𝑡
= 5000 joule/ gram ℃
1000 W x 90 sec. = 90000 J is ∆𝑄 4
∆𝑄 = 𝑚 𝑥 𝑐 𝑥 ∆𝑡
c=
90000
0.5 𝑥 50
so c =
∆𝑄
𝑚 𝑥 ∆𝑡
= 4800 joule/ gram ℃
Gözde ÇEVİK
Table 3: The specific heat capacity for 4 trials and the mean of them
TRIAL NO:
Specific Heat Capacity
( joule/ gram ℃)
1st
6000
2nd
5000
3rd
5000
4th
4800
The Mean
5200
CONCLUSION AND EVALUATION
At the end of this experiment, the specific heat capacity of water is found 5200
joule/ gram℃. However, specific heat capacity of water is 4.186 joule/gram °C according
to the physics data booklet. The value that’s found at the end of this experiment is much
bigger than the actual value. This may be caused because of the heat loss to the
surroundings during the experiment and using isolation may prevent this loss. Using more
precise thermometer is also an improvement, especially a digital thermometer to prevent
the false measurements.