Specific Heat Capacity of Water (DCP and CE)

Topics: Heat, Thermodynamics, Specific heat capacity Pages: 8 (1216 words) Published: October 22, 2013
Aim:To determine the specific heat capacity of water by heating water and recording temperatures at regular intervals. Hypothesis:(i) P = Q /t
(ii) Q = m c ΔT
Where,
P is the power of the heater
Q is the total energy provided by the heater
t is the total time for which the heater was used
m is the mass of water
c is the specific heat capacity of water, and
ΔT is the change in temperature
From (i) and (ii)m cΔT = P t
ΔT = (P/mc) * t
This is in the form y = mx + c, where
y = ΔT
m = (P/mc), and
x = t
Therefore, plotting ΔT in the y-axis with t in the x-axis gives us the slope (P/mc), from which we can find ‘c’.

DATA COLLECTION AND PROCESSING
Volume of water = 400 ± 1 cm3?[measured using a beaker of least count 1 cm3] Density of water = 1 gcm-3
Therefore, Mass of water = Density x Volume = 400 g = 0.400 kg %error in mass = % error in volume = 1/400 * 100 = 0.3%
Error in mass = 0.3% of 0.400 kg = 0.001 kg
Mass = 0.400 ± 0.001 kg
Power of heater = 500W

Table 1: Raw data table showing temperatures at intervals of 15 seconds while heating Time interval/s ± 0.1 s
Temperature/°C
± 0.5°C

i
Ii
Iii
iv
15.0
27.0
28.0
28.0
27.0
30.0
33.0
33.0
34.0
34.0
45.0
38.0
39.0
39.0
39.0
60.0
43.0
44.0
43.0
44.0
75.0
48.0
48.0
48.0
50.0
90.0
51.0
52.0
53.0
54.0
105.0
56.0
58.0
58.0
57.0
120.0
61.0
62.0
62.0
62.0
135.0
66.0
67.0
67.0
67.0
The uncertainty in time interval is taken 0.1s because the stopwatch we used had a least count of 0.1 s and was digital. Here, uncertainty in temperature is 0.5°C because the least count of the thermometer was 1°C and it was an analog thermometer. So uncertainty is half the least count. Now, the average temperature was calculated by taking the sum of the four temperature values for a specific time and dividing the sum by 4. The uncertainty in temperature was calculated by finding the difference between the maximum value and the minimum value and then dividing it by 2. Sample Calculations:

1. Average temperature = (27+28+28+27)/4 = 27.5 = 28°C
2. Uncertainty in temperature = (Max-Min)/2 = (28-27)/2 = 0.5 = 1°C Table 2: Table showing average temperature and uncertainty in this average with corresponding time Time/s ± 0.1 s
Average temperature/°C
Uncertainty in temperature/°C

15.0
27.5
1
30.0
33.5
1
45.0
38.8
1
60.0
43.5
1
75.0
48.5
1
90.0
52.5
2
105.0
57.3
1
120.0
61.8
1
135.0
66.8
1
From the above table, time is graphed on the x-axis, average temperature on the y-axis and the uncertainty column is used for error bars for y-axis.

For max gradient:For min gradient:
Time/s
Temperature/°C
30
32.5
135
67.8
Time/s
Temperature/°C
30
34.5
135
65.8

Graph 1: Graph showing temperature against time

Note: The first point (15, 27.5) is an outlier in the graph as it only touches the max gradient line and not the others. Also, X-error bars too small to be visible (0.1s)

Scale:
x-axis 1 unit = 10 sec
y-axis 1 unit = 2 °C
y-intercept = 23.79 °C [ranging from 22.41 °C to 25.56 °C) Slope = 0.3201 °C s-1
Uncertainty in slope= (Max slope – Min slope)/2
= (0.3362 – 0.2981)/2
= 0.02 °C s-1
%uncertainty in slope = 0.01/0.3201*100 = 3%
Now,
P/mc = 0.3201
c = P/(m*0.3201)
= 500/(0.400*0.3201)
= 3905.03
= 3900 J kg-1°C-1
For % uncertainty,
%error in mass = % error in volume = 1/400 * 100 = 0.3%
The value of the power given in the packaging of the heater is considered to be very precise with negligible error. So,
% uncertainty = (0.3 + 3) % = 3.3% = 3%
Uncertainty = 3/100 * 3900 = 117 = 100 J kg-1°C-1
Therefore,
Specific heat capacity of water = 3900 ± 100 J kg-1 °C-1

CONCLUSION AND EVALUATION
Specific heat capacity is the amount of heat energy required to raise the temperature of a body by 1°C per unit of mass. http://chemistry.about.com/od/chemistryglossary/g/Specific-Heat-Capacity-Definition.html Here,

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