In the previous section we saw the details about the Kelvin scale to measure temperatures. In this section we will see Specific heat capacity.
Some experiments related to heat was conducted on various substances like Coconut oil, Copper, Lead etc., The observations can be written as follows:
Experiment 1:
• 10 kg of coconut oil was taken.
• It's initial temperature in K was noted down.
• Then it was heated until the temperature increased by 20 K. (For example, if the initial temperature was 301 K, the heating was stopped when the temperature became 321 K)
• The amount of heat energy required to bring about this increase in temperature was found to be 420000 joules
Experiment 2:
• 10 kg of copper was taken.
• It's initial temperature in K was noted down.
• Then it was heated until the temperature increased by 20 K.
• The amount of heat energy required to bring about this increase in temperature was found to be 77000 joules
Experiment 3:
• 10 kg of water was taken.
• It's initial temperature in K was noted down.
• Then it was heated until the temperature increased by 20 K.
• The amount of heat energy required to bring about this increase in temperature was found to be 837200 joules
Experiment 4:
• 10 kg of lead was taken.
• It's initial temperature in K was noted down.
• Then it was heated until the temperature increased by 20 K.
• The amount of heat energy required to bring about this increase in temperature was found to be 24000 joules
Now we can do some calculations:
CALCULATION A:
In experiment 1:
1. The heat energy required to bring about a rise of 20 K is 420000 J
2. So the heat energy required to bring about a rise of 1 K = 420000⁄20 = 210000 J
In experiment 2:
1. The heat energy required to bring about a rise of 20 K is 77000 J
2. So the heat energy required to bring about a rise of 1 K = 77000⁄20 = 3850 J
In experiment 3:
1. The heat energy required to bring about a rise of 20 K is 1260000 J
2. So the heat energy required to bring about a rise of 1 K = 837200⁄20 = 41860 J
In experiment 4:
1. The heat energy required to bring about a rise of 20 K is 24000 J
2. So the heat energy required to bring about a rise of 1 K = 24000⁄20 = 1200 J
So we can write:
• A sample of 10 kg coconut oil will require a heat energy of 210000 for increasing it's temperature by 1 K
• A sample of 10 kg copper will require a heat energy of 3850 J for increasing it's temperature by 1 K
• A sample of 10 kg water will require a heat energy of 63000 J for increasing it's temperature by 1 K
• A sample of 10 kg lead will require a heat energy of 1200 J for increasing it's temperature by 1 K
■ The heat energy required to raise the temperature of a sample of a substance by 1 K is called the heat capacity of that sample. It's unit is J/K or J K -1
• So we can write:
• Heat capacity of a sample of 10 Kg of coconut oil is 210000 J K -1
• Heat capacity of a sample of 10 Kg of copper is 3850 J K -1
• Heat capacity of a sample of 10 Kg of water is 41860 J K -1
• Heat capacity of a sample of 10 Kg of lead is 1200 J K -1
Note the use of the word 'sample' in the above steps. A sample can have any mass: 10 kg, 12 kg, 12.5 kg . . . So, to standardise, we will take unit mass. That is., 1 kg.
CALCULATION B:
1. We have: Heat capacity of a sample of 10 Kg of coconut oil is 210000 J K -1
• So heat capacity of 1 kg of coconut oil = 210000⁄10 = 21000 J
2. We have: Heat capacity of a sample of 10 Kg of copper is 3850 J K -1
• So heat capacity of 1 kg of copper = 3850⁄10 = 385 J
3. We have: Heat capacity of a sample of 10 Kg of water is 63000 J K -1
• So heat capacity of 1 kg of water = 41860⁄10 = 4186 J
4. We have: Heat capacity of a sample of 10 Kg of lead is 1200 J K -1
• So heat capacity of 1 kg of lead = 1200⁄10 = 120 J
■ The heat capacity of 1 kg of a substance is called the Specific heat capacity of that substance
Specific heat capacity (c) is a constant for a substance. That is., every substance has it's own unique value for 'c'. The following table 6.1 shows the c values of some common substances:
From the table, we can see that water has a high specific heat capacity than others It's value is 4186 J kg-1 K-1. For use in numerical problems, it is assumed as 4200 J kg-1 K-1.
Solved example 6.5
Calculate the quantity of heat required to raise the temperature of 5 kg of iron from 303 K to 343 K. Specific heat capacity of iron is 460 J kg-1 K-1
Solution:
1. We have : Q = mcθ.
2. θ = (θ2 - θ1) = (343 - 303) = 40 K
3. Substituting the known values we get: Q = 5×460×40 = 92000 J
Solved example 6.6
The temperature of 0.5 kg of water is 303 K. It is cooled to 278 K by keeping it in a refrigerator. What is the time taken by the water to reach 278 K, if 87.5 J of heat is given out in each second? Specific heat capacity of water is 4200.
Solution:
1. We have : Q = mcθ.
2. θ = (θ2 - θ1) = (278 - 303) = -25 K
• The negative sign indicates that, heat energy is given out
3. Substituting the known values we get: Q = 0.5×4200×-25 = -52500 J
4. Energy given out per second = -87.5 J
5. So time taken = -52500⁄-87.5 = 600 seconds = 10 minutes
Solved example 6.7
Heat required to increase the temperature of 10 kg of water by 1 K is 42000 J. Then how much heat will be required to increase the temperature of 1 kg of water by 1 K?
Solution:
1. We have : Q = mcθ.
2. Substituting the known values we get: 42000 = 10×c×1 ⟹ 42000 = 10c ⟹ c = 4200 J kg-1 K-1
3. In part 2, we have: m = 1 kg, c = 4200 J kg-1 K-1 (calculated in (2) above), θ= 1 K
4. Substituting in (1) we get: Q = 1×4200×1 = 4200 J
Some experiments related to heat was conducted on various substances like Coconut oil, Copper, Lead etc., The observations can be written as follows:
Experiment 1:
• 10 kg of coconut oil was taken.
• It's initial temperature in K was noted down.
• Then it was heated until the temperature increased by 20 K. (For example, if the initial temperature was 301 K, the heating was stopped when the temperature became 321 K)
• The amount of heat energy required to bring about this increase in temperature was found to be 420000 joules
Experiment 2:
• 10 kg of copper was taken.
• It's initial temperature in K was noted down.
• Then it was heated until the temperature increased by 20 K.
• The amount of heat energy required to bring about this increase in temperature was found to be 77000 joules
Experiment 3:
• 10 kg of water was taken.
• It's initial temperature in K was noted down.
• Then it was heated until the temperature increased by 20 K.
• The amount of heat energy required to bring about this increase in temperature was found to be 837200 joules
Experiment 4:
• 10 kg of lead was taken.
• It's initial temperature in K was noted down.
• Then it was heated until the temperature increased by 20 K.
• The amount of heat energy required to bring about this increase in temperature was found to be 24000 joules
Now we can do some calculations:
CALCULATION A:
In experiment 1:
1. The heat energy required to bring about a rise of 20 K is 420000 J
2. So the heat energy required to bring about a rise of 1 K = 420000⁄20 = 210000 J
In experiment 2:
1. The heat energy required to bring about a rise of 20 K is 77000 J
2. So the heat energy required to bring about a rise of 1 K = 77000⁄20 = 3850 J
In experiment 3:
1. The heat energy required to bring about a rise of 20 K is 1260000 J
2. So the heat energy required to bring about a rise of 1 K = 837200⁄20 = 41860 J
In experiment 4:
1. The heat energy required to bring about a rise of 20 K is 24000 J
2. So the heat energy required to bring about a rise of 1 K = 24000⁄20 = 1200 J
So we can write:
• A sample of 10 kg coconut oil will require a heat energy of 210000 for increasing it's temperature by 1 K
• A sample of 10 kg copper will require a heat energy of 3850 J for increasing it's temperature by 1 K
• A sample of 10 kg water will require a heat energy of 63000 J for increasing it's temperature by 1 K
• A sample of 10 kg lead will require a heat energy of 1200 J for increasing it's temperature by 1 K
■ The heat energy required to raise the temperature of a sample of a substance by 1 K is called the heat capacity of that sample. It's unit is J/K or J K -1
• So we can write:
• Heat capacity of a sample of 10 Kg of coconut oil is 210000 J K -1
• Heat capacity of a sample of 10 Kg of copper is 3850 J K -1
• Heat capacity of a sample of 10 Kg of water is 41860 J K -1
• Heat capacity of a sample of 10 Kg of lead is 1200 J K -1
Note the use of the word 'sample' in the above steps. A sample can have any mass: 10 kg, 12 kg, 12.5 kg . . . So, to standardise, we will take unit mass. That is., 1 kg.
CALCULATION B:
1. We have: Heat capacity of a sample of 10 Kg of coconut oil is 210000 J K -1
• So heat capacity of 1 kg of coconut oil = 210000⁄10 = 21000 J
2. We have: Heat capacity of a sample of 10 Kg of copper is 3850 J K -1
• So heat capacity of 1 kg of copper = 3850⁄10 = 385 J
3. We have: Heat capacity of a sample of 10 Kg of water is 63000 J K -1
• So heat capacity of 1 kg of water = 41860⁄10 = 4186 J
4. We have: Heat capacity of a sample of 10 Kg of lead is 1200 J K -1
• So heat capacity of 1 kg of lead = 1200⁄10 = 120 J
In the above calculations, we calculated the specific heat capacity of some common substances: Coconut oil, copper, water and lead. We can now write the general procedure to find the specific heat capacity of any given substance.
1. Take a sample of the given substance. Let the mass of the sample be 'm' kg
2. Note down the initial temperature θ1 in K
3. Heat the sample for some time. Note down the final temperature. Let this be θ2.
• Then the rise in temperature = (θ2 - θ1). Let this be 'θ' K
• Then the rise in temperature = (θ2 - θ1). Let this be 'θ' K
4. Note down the heat energy required to bring about the rise in temperature of 'θ' K. Let this energy be 'Q' joules
5. Then we get:
• Heat capacity of the sample
= The heat energy required to raise the temperature of a sample of a substance by 1 K
= Q⁄θ
= The heat energy required to raise the temperature of a sample of a substance by 1 K
= Q⁄θ
• Specific heat capacity of the substance in the sample
= The heat capacity of 1 kg of a substance
= Q⁄θ×m
6. The 'specific heat capacity' is denoted by the symbol 'c'. So we can write:
Eq.6.6:
c = Q⁄θ×m .
7. Now we must write the units of 'c':
• In Eq.6.6 above, we have:
♦ Energy (whose unit is joules) in the numerator
♦ Product of temperature (whose unit is K) and mass (whose unit is kg) in the denominator
• So the unit of 'c' is: J⁄K×kg = J kg-1 K-1. (joules per kilo grams per kelvin)
■ So we can write the definition for Specific heat capacity:
Specific heat capacity of a substance is the heat energy required to raise the temperature of 1 kg of that substance through 1 K
Specific heat capacity (c) is a constant for a substance. That is., every substance has it's own unique value for 'c'. The following table 6.1 shows the c values of some common substances:
Table 6.1
| Substance | Specific heat capacity (J kg-1 K-1) |
|---|---|
| Water | 4186 |
| Ice | 2130 |
| Steam | 460 |
| Sea water | 3900 |
| Glass | 500 |
| Iron | 460 |
| Copper | 385 |
| Silver | 234 |
| Lead | 120 |
• The high specific heat capacity of water implies that, lot of heat energy is required to make 'even small increases' in the temperature of water.
• Now we will see some situations in our day to day life, where this high 'c' value of water have much significance:
1. Water in the rivers and ponds do not get heated even if the temperature of the surrounding environment increases on a hot sunny day. This makes aquatic life possible
2. Body of plants and animals are about 80% water. So the body temperature do not increase rapidly even if the temperature of the surrounding environment increases on a hot sunny day.
3. In motor vehicles, large quantities of heat are produced by the engines continuously. This heat is transmitted to water which is used as a coolant. Because of the high 'c' value, water can absorb large quantities of heat with out itself getting much hot.
4. We know that large quantities of heat are absorbed by water (for a small increase in temperature) when it is heated by an external source
• The reverse happens when it is cooled. That is., even for a small decrease in temperature, large quantities of heat will have to be released.
• If there is no external cooling device, this fall in temperature will happen only slowly.
• So water, once heated, will remain hot for a long time. This property makes it useful in 'hot water bags', which are used in the treatments of some illness.
5. Land breeze and Sea breeze
• The 'c' value of sand is only 1/5 of water. So it is easier to heat up the land.
• During day time, the land quickly gets hotter than sea. So the air above land will be hotter than the air above sea.
• The hot air will rise up creating a void space near the surface of the land. The cold air above sea will then blows towards this void. This is sea breeze.
• During night, the reverse happens. The land will quickly become cold.
• But sea will become cold only slowly. So the air above sea will be hotter than the air above land.
• The hot air will rise up creating a void space near the surface of the sea. The cold air above land will then blows towards this void. This is land breeze.
Now we will see a practical application of specific heat capacity
1. Consider a sample of a substance.
2. Let the specific heat capacity of the substance be 'c'
• Then the heat energy required to raise the temperature of 1 kg of that substance through 1 K = c joules
3. Let the mass of the sample be m kg
• Then the heat energy required to raise the temperature of that whole sample through 1 K = mc joules
• Then the heat energy required to raise the temperature of that whole sample through θ K = mcθ joules
■ So we can write:
If Q is the total heat energy required to raise the temperature of 'm' kg of a substance (whose specific heat capacity is 'c') through theta K, Then
Eq.6.7
Q = mcθ joules
Let us see some solved examples:
Solved example 6.4
(a) c of copper is 385 J kg-1 K-1. How much heat energy is required to raise the temperature of 1 kg of copper by 10 K
(b) c of iron is 460 J kg-1 K-1. How much heat energy is required to raise the temperature of 1 kg of copper by 20 K
(c) c of water is 4200 J kg-1 K-1. How much in K will the temperature of 2 kg water rise, if it receives 42000 J of heat energy
(d) When 1 kg of lead received 1200 J of heat energy, it's temperature was raised by 10 K. What is the 'c' of lead?
Solution:
Part (a):
1. We have : Q = mcθ.
2. Substituting the known values we get: Q = 1×385×10 = 3850 J
Part (b):
1. We have : Q = mcθ.
2. Substituting the known values we get: Q = 1×460×20 = 9200 J
Part (c):
1. We have : Q = mcθ.
2. Substituting the known values we get: 42000 = 2×4200×θ ⟹ 10 = 2×θ ⟹ θ = 5 K
Part (d):
1. We have : Q = mcθ.
2. Substituting the known values we get: 1200 = 1×c×10 ⟹ 1200 = 10c ⟹ c = 120 J kg-1 K-1
Solved example 6.5
Solution:
1. We have : Q = mcθ.
2. θ = (θ2 - θ1) = (343 - 303) = 40 K
3. Substituting the known values we get: Q = 5×460×40 = 92000 J
Solved example 6.6
The temperature of 0.5 kg of water is 303 K. It is cooled to 278 K by keeping it in a refrigerator. What is the time taken by the water to reach 278 K, if 87.5 J of heat is given out in each second? Specific heat capacity of water is 4200.
Solution:
1. We have : Q = mcθ.
2. θ = (θ2 - θ1) = (278 - 303) = -25 K
• The negative sign indicates that, heat energy is given out
3. Substituting the known values we get: Q = 0.5×4200×-25 = -52500 J
4. Energy given out per second = -87.5 J
5. So time taken = -52500⁄-87.5 = 600 seconds = 10 minutes
Solved example 6.7
Heat required to increase the temperature of 10 kg of water by 1 K is 42000 J. Then how much heat will be required to increase the temperature of 1 kg of water by 1 K?
Solution:
1. We have : Q = mcθ.
2. Substituting the known values we get: 42000 = 10×c×1 ⟹ 42000 = 10c ⟹ c = 4200 J kg-1 K-1
3. In part 2, we have: m = 1 kg, c = 4200 J kg-1 K-1 (calculated in (2) above), θ= 1 K
4. Substituting in (1) we get: Q = 1×4200×1 = 4200 J
No comments:
Post a Comment