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What are the Advantages of Calorimetry?
Fundamental Principle of Calorimetry
When two bodies at different temperatures are brought in contact, heat is transferred from the body at higher temperature to that at lower temperature. So, the former starts to cool down whereas the latter starts to warm up. The flow of heat continues till both reach the same temperature. The state where both the bodies are at the same temperature is called thermal equilibrium.
Now we know that heat is a form of energy. According to the law of conservation of energy, energy can neither be created nor be destroyed. So during such heat transfers, if heat does not transform into another form of energy (or, if no other form of energy transforms into heat energy) then, heat released by a body = heat absorbed by another body This is the fundamental principle of calorimetry. This is also true for any number of bodies brought in contact with one another. In that case, we can write down the principle of calorimetry as:
heat released by the hot bodies = heat absorbed by the cold bodies.
Calorimeter
A calorimeter is a cylindrical metallic container (C) mostly used in calorimetric experiments [Fig.] which is generally made of copper. The container is partly filled with a proper liquid called calorimetric substance. A stirrer (S) made of copper is used to stir the liquid and a thermometer (T) is used to measure the temperature of the liquid. Both are immersed in the liquid.
In a calorimetric experiment when an object is dipped into the liquid of the calorimeter there must be a difference in temperature between the object and the liquid. After some time the object, the calorimeter and the liquid inside the calorimeter reach thermal equilibrium.
During this interval of time if
- no heat exchange occurs between the calorimeter and the surrounding,
- no exothermic or endothermic reaction (physical or chemical) happens inside the calorimeter and
- the object does not dissolve in the liquid, then according to the fundamental principle of calorimetry, heat released by the hot objects = heat absorbed by the cold objects.
There will be error in the experimental result if the above three conditions are not satisfied. Heat exchange may take place between the calorimeter and the environment by means of—
- conduction,
- convection and
- radiation.
A calorimeter is so constructed that heat exchange by any of the three processes can be minimised.
Calorimetric Substance
Voter:
i) Advantages:
- For the same rise or fall in temperature, heat gained or lost by water is more than that gained or lost by the same mass of other solids or liquids of lower specific heat.
- Water is commonly used as a calorimetric substance because water is easily available.
ii) Disadvantages:
1. Again due to high specific heat of water, for absorption of same amount of heat, rise in temperature of water is less than that of other liquids of lower specific heat. Now if the increase of temperature is low then there is high chance of error in reading.
2. The boiling point of water is 100°C. If an object of tem-perature more than 100°C is mixed with water, a sudden vaporisation occurs.
Due to these reasons water is not useful as a calorimetric substance.
Aniline: Recently aniline has been recognised as the best calorimetric substance. Aniline is present in pure form. Its specific heat is low (0.62 cal ᐧ g-1 ᐧ °C-1) and boiling point is high (183.9°C).
Determination of High Temperature
A mercury thermometer cannot be used directly for the measurement of temperature of a furnace or a flame. But the fundamental principle of calorimetry may be adopted to measure such high temperatures accurately and easily. A small piece of solid whose specific heat is known and which has a sufficiently high melting point (e.g., iron whose melting point is 1811 K) is to be placed in the source for a considerable period of time and allowed to attain thermal equilibrium with the source. It is then quickly transferred to the water in a calorimeter.
Initial temperature of the water is to be noted before the addition of the solid. The water is stirred well and the final temperature attained is measured by a thermometer. There should be sufficient water in the calorimeter so that the final temperature attained remains below the upper fixed point of the thermometer used.
Calculation: Let the temperature of the source = t2 = initial temperature of the solid, mass of the solid taken = M, specific heat of the solid = s, water equivalent of the calorimeter with stirrer = W, mass of the’water in the calorimeter = m, its specific heat = sw, initial temperature of the calorimeter and the water = t1, the final temperature of the water and the solid = t.
∴ Heat lost by the solid = Ms(t2 – t)
Heat gained by the calorimeter and its stirrer = Wsw(t – t1)
Heat gained by the water = m × sw × (t – t1)
Hence, from the fundamental principle of calorimetry,
As magnitudes of all the quantities in the right hand side of equation (1) are known, t2 can be calculated.
Numerical Examples
Example 1.
50 g of an alloy containing 80% copper and 20% silver is heated up to t1 = 80°C and then dropped in a calorimeter of water equivalent W = 10 g containing t2 = 90 g of water at t2 = 20°C. What will be the final temperature of the mixture? Specific heat capacity of copper and silver are sc = 0.09 cal ᐧ g-1 ᐧ °C-1 and ss = 0.15 cal ᐧ g-1 ᐧ °C-1 respectively.
Solution:
Mass of copper in the alloy, mc = \(\frac{50 \times 80}{100}\) = 40 g
∴ Mass of silver, ms = 50 – 40 = 10 g
Let the final temperature of the mixture be t.
Applying the formula, H = ms t
Heat lost by copper = 40 × 0.09 × (80 – t) = (288 – 3.61) cal
Heat lost by silver = 10 × 0.05 × (80 – t) = (40 – 0.51) cal
Total heat lost by copper and silver = (328 – 4.1 t) cal
Heat gained by water and the calorimeter
= (10 + 90) × 1 × (t – 20) = (100t – 2000) cal
∴ Heat lost = heat gained
∴ 328 – 4.1 t = 100t – 2000
or t = \(\frac{2328}{104.1}\) = 22.36°C.
∴ Final temperature of the mixutre is 22.36°C.
Example 2.
A piece of platinum of mass 200 g, is heated for a sufficiently long time in a furnace and dropped quickly in 650 g of water at 10°C, kept in a container of water equivalent 50g. Final temperature of the mixture becomes 25°C. If the specific heat capacity of platinum is 0.03 cal ᐧ g-1 ᐧ °C-1, find the temperature of the furnace.
Solution:
Let the temperature of the furnace be t°C which is also the initial temperature of the piece of platinum.
Heat lost by the platinum piece
= 200 × 0.03 × (t – 25) = 6(t – 25) cal
Heat gained by the container and water
= 50 × (23 – 10) + 650 × 1 × (25 – 10)
= 750 + 9750 = 10500 cal
Assuming no loss of heat to the surroundings, heat lost = heat gained
or, 6(t – 25) = 10500 or, t = 1775°C
∴ Temperature of the furnace is 1775°C.
Example 3.
A solid of mass 70 g is heated and dropped in a cal-orimeter of water equivalent 10 g containing 116 g of water. If the fall in temperature of the solid is 15 times the rise in temperature of water, find the specific heat capacity of the solid.
Solution:
Let t be the rise in temperature of water and s then specific heat of the solid.
Fall in temperature of the solid = 15 r Heat lost by the solid = 70 × s × 15t
Heat gained by the calorimeter and water
= 10 × t + 116 × 1 × t = 126t
Since, heat lost = heat gained
70 × s × 15t = 126t
or, s = \(\frac{126}{70 \times 15}\) = 0.12 cal ᐧ g-1 ᐧ C-1
Example 4.
Ratio between the densities and specific heats of the materials of two bodies are 2:3 and 0.12:0.09 respectively. If their volumes are in the ratio 7:8, what is the ratio of their thermal capacities?
Solution:
Let the density of the first substance be d1, its specific heat s1 and volume V1.
Also, let the density, specific heat and volume for the second body be d2, s2 and V2 respectively.
Hence, their thermal capacities are in the ratio of 7 : 9.
Example 5.
A body of mass 100 g is heated up to 122°C and dropped quickly in water of mass 300 g kept at 28°C in a copper calorimeter of mass 50 g. The final temperature of the mixture becomes 30 °C. If the specific heat of copper is 0.09 cal ᐧ g-1 ᐧ °C-1, find the specific heat of the material of the body.
Solution:
Let the specific heat of the material be s. Heat lost by the body = 100 × s × (122 – 30)
Heat lost by the body = 100 × s × 92 = 9200 s cal
Heat gained by water = 300 × 1 × (30 – 28)
= 300 × 2 = 600 cal
Heat gained by the calorimeter
= 50 × 0.09 × (30 – 28) = 4.5 × 2 = 9 cal
We know, heat lost = heat gained
or, 9200s = 600 + 9
or, s = \(\frac{609}{9200}\) = 0.0662 cal ᐧ g-1 ᐧ °C-1
Example 6.
Calculate the increase in energy (in joule) per atom of a piece of aluminium when its temperature is raised by 1°C. Given, 27 g of aluminium contains 6 × 1023 atoms, and specific heat capacity of aluminium = 0.2 cal ᐧ g-1 ᐧ °C-1.
Solution:
Heat required to raise the temperature of 1 g alu-minium through 1°C = 1 × 0.2 × 1 = 0.2 cal = 0.2 × 4.2 J = 0.84 J.
Number of atoms in 1 g of aluminium = \(\frac{6 \times 10^{23}}{27}\)
Hence, increase in energy per atom = \(\frac{0.84}{\frac{6 \times 10^{23}}{27}}\)J
= 3.78 × 10-23J
Example 7.
When 210 g of water at 80°C is kept in a calorimeter of water equivalent 90 g, its temperature decreases to 60°C in 10 min. If the water is replaced by a liquid of mass 100 g, the same fall in temperature takes place in 5 min. If the rate of cooling is the same in both cases, what is the specific heat capacity of the liquid?
Solution:
Heat lost by the calorimeter and water in 10 min = 90 × (80 – 60) + 210 × 1 × (80 – 60) = 6000 cal
∴ Rate of cooling = \(\frac{6000 \mathrm{cal}}{10 \mathrm{~min}}\) = 600 cal ᐧ min-1
Let the specific heat of the liquid be s.
Heat lost by the calorimeter and liquid in 5 min
= 90 × (80 – 60) + 100 × s × (80 – 60)
= (1800 + 2000s) cal
Hence, rate of cooling = \(\frac{(1800+2000 s)}{5}\) cal ᐧ min-1
As per given condition,
\(\frac{(1800+2000 s)}{5}\) = 600
or, 1800 + 2000s = 3000
or, s = 0.6 cal ᐧ g-1 ᐧ °C-1
Example 8.
Liquids A, B and C are at temperatures 15°C, 25 °C and 35 °C respectively. When equal masses of A and B are mixed, the temperature becomes 21 °C. Temperature of the mixture of equal masses of B and C becomes 32°C.
(i) Show that the ratio of the specific heats of A and C is 2 : 7.
(ii) What will be the final temperature of the mixture of equal masses of A and C?
Solution:
Let the specific heats of A, B and C be sA, sB and sC respectively.
i) When liquid A of mass m and liquid B of mass m are mixed, heat lost by B = heat gained by A i.e., m × sA × (25 – 21) = m × sA × (21 – 15) or,
or, sA ᐧ 6 = sB ᐧ 4
or, \(\frac{s_A}{s_B}\) = \(\frac{2}{3}\) ……. (1)
When liquid B of mass m and liquid C of mass m are mixed, heat lost by C = heat gained by B.
i.e., m × sC × (35 – 32) = m × sB × (32 – 25) or,
or, \(\frac{s_B}{s_C}\) = \(\frac{3}{7}\) ……. (2)
From (1) and (2), we get
\(\frac{s_A}{s_B}\) × \(\frac{s_B}{s_C}\) = \(\frac{2}{3}\) × \(\frac{3}{7}\) or, \(\frac{s_A}{s_C}\) = \(\frac{35-t}{t-15}\) or, \(\frac{2}{7}\) = \(\frac{35-t}{t-15}\) or, t = 30.56°C
∴ Temperature of the mixture is 30.56°C
Example 9.
An iron ball of mass 10 g and of specific heat 0.1 cal g-1 ᐧ °C-1 is heated in a furnace and quickly transferred to a thick-walled copper vessel of mass 200 g and of specific heat 0.09 cal . g-1 ᐧ °C-1, kept at 50°C. The vessel along with its contents, is placed in a calorimeter of water equivalent 20g, containing 180 g of water at 20°C. A thermometer, dipped in the water of the calorimeter shows the maximum temperature to be 26°C. Find the temperature of the furnace. Will there be any local boiling of water in the calorimeter?
Solution:
Let the temperature of the furnace be θ.
Heat lost by the hot iron ball and copper vessel
= 10 × 0.1 × (θ – 26) + 200 × 0.09 × (50 – 26)
= (θ – 26) + 18 × 24 = (θ + 406) cal
Heat gained by the calorimeter and water
=(20 + 180) × (26 – 20) = 1200 cal
∵ Heat gain = heat loss
∵ θ + 406 = 1200 or, θ = 1200 – 406 = 794°C
∴ Temperature of the furnace is 794°C.
Let the intermediate temperature attained by the iron ball and copper vessel be t.
As heat lost by the iron ball = heat gained by the copper vessel
So, 10 × 0.1 × (794 – t) = 200 × 0.09 × (t – 50)
or, 794 – t = 18t – 900
∴ t = 89.16°C
∴ Temperature of the system of iron ball and copper vessel is 89.16°C. This is below the boiling point 100°C of water. So there will be no local boiling in the calorimeter water.
Example 10.
Three liquids of the same amount are mixed. The specific heats of these liquids are s1, s2 and s3 and their initial temperatures are θ1, θ2 and θ3 respectively. Find out the final temperature of the mixture. [Hs ‘05]
Solution:
Let the final temperature of the mixture be θ and the mass of each liquid be m.
∴ Heat absorbed by the 1st liquid, H1 = ms1(θ – θ1)
Heat absorbed by the 2nd liquid, H2 = ms2(θ – θ2)
Heat absorbed by the 3rd liquid, H3 = ms3(θ – θ3)
Since no heat is supplied from outside,
∴ H1 + H2 + H3 = 0 ……… (1)
[any one or two of H1, H2, H3 must be negative, implying heat lost; then equation (1) means, heat gained = heat lost]
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