Physics Topics can be challenging to grasp, but the rewards for understanding them are immense.
What is the Unit of Heat? What are the Disadvantages of Water’s High Specific Heat?
Introduction
An important characteristic of heat must be noted from the very beginning. Unlike the quantities volume, pressure, temperature etc., heat is not related to any equilibrium state of a body. Heat manifests itself only when a body undergoes a transition from one state to another. Statements like ‘heat of a body’, ‘initial heat’ and ‘final heat’ have no physical meaning. Physically significant are the quantities like ‘heat gained’, or heat lost’ by a body during any process. In this sense, heat is regarded as an energy in transit.
The rise or fall in temperature of a body implies addition or extraction of heat from it. If two objects are in contact with each other then there will be heat flow between them till both attain an equilibrium temperature. In such situations a need arises to know the quantity of heat transferred from one body to another, or from an object to its surroundings. The study of problems that involve heat exchange between different bodies is called calorimetry.
Factors Affecting Heat Transfer
We know that it takes longer to boil some water than to warm it. Also boiling water takes longer to cool down than lukewarm water does. So for rise and fall of different amount of temperature, different amounts of heat are absorbed or released.
Again, increasing the amount of water to be boiled, increases the time taken to boil it too. From this we can conclude that the amounts of heat absorbed (or released) by the bodies of same material but different mass for a fixed rise (or fall) in temperature depend on their masses.
When a piece of iron and some water of the same mass are heated, the piece of iron heats up faster than the water.
Clearly, the material of an object plays a role in determining the amount of heat absorbed (or released) by the object for a certain rise (or fall) in temperature.
From these observations, it can be inferred that, the amount of heat absorbed or released by a body depends on the change of temperature of the body, mass of the body and material of the body, provided that the material does not undergo any phase change.
- Temperature: Amount of heat gained or lost by a body of fixed mass is directly proportional to the rise or fall in its temperature.
- Mass: Amount of heat gained or lost by a body, for a fixed rise or fall in temperature, is directly proportional to its mass.
- Material: Amount of heat gained or lost by a body depends on the material of the body.
When a body of mass m receives heat H, so that the rise in temperature is t,
H ∝ mt or, H = smt …… (1)
where s is specific heat whose value depends on the material of the body.
Therefore, H (the heat absorbed or released by a body) = m (mass of the body) × s (specific heat of the material of the body) × t (rise or fall in temperature)
It is to be noted that t in equation (1) is the change in temperature of the body. So only the heat gained or lost can be measured by this equation.
Units of Heat
Some useful units for the measurement of heat are defined as follows:
Calorie: It is the unit of heat in CGS system. The amount of heat required to raise the temperature of 1 g of water from 14.5 °C to 15.5 °C is called 1 calorie (1 cal).
Practically, the heat required to raise the temperature of 1 g of pure water from 0°C to 1°C is not the same as the heat required to raise its temperature, for example, from 45°C to 46°C, though the rise in temperature is the same. Hence, quantity of heat differs at the different ranges in the scale of temperature. So, most acceptable definition of calorie is,
1 cal = \(\frac{1}{100}\) × heat required to raise the temperature of 1 g of pure water from 0°C to 100°C. It is often known as mean calorie. Its value is equal to the amount of heat required to raise the temperature of 1 g of pure water from 14.5° C to 15.5°C. So it is sometimes called 15°C calorie.
Kilocalorie or kilogram calorie: Quantity of heat required to raise the temperature of 1 kg of pure water from 14.5°C to 15.5°C, is called 1 kilocalorie.
As 1 kg = 1000 g, 1 kilocalorie (kcal) = 1000 calorie (cal).
Joule: This is the unit of heat in SI. 1 cal = 4.2 J.
Specific Heat or Specific Heat Capacity
In section 7.2, it has been stated that heat gained (or lost) by a body depends on its mass, the rise (or fall) in temperature and the material of the body.
Let us take two bodies of same mass but made of different materials. If we try to change the temperature of both the bodies by the same amount, we will find that the heat required is different for them. Therefore it is obvious, the quantity of heat required also depends on a special property.
This property is known as specific heat or specific heat capacity of the body. It depends on the material of the body. It is obvious that different materials possess different specific heat capacities.
Definition: Specific heat capacity is the quantity of heat required to raise the temperature of unit mass of a substance by unit amount.
Definition of specific heat capacity in CGS system:
It is the quantity of heat required to raise the temperature of 1 g of a substance through 1 °C. In this system the unit of specific heat is cal ᐧ g-1 ᐧ °C.
For example, ‘the specific heat capacity of copper is 0.093 cal. g-1 ᐧ °C-1 means that to raise the temperature of 1 g of copper by 1 °C, the heat required is 0.093 cal.
Definition of specific heat capacity in SI : The quantity of heat required to raise the temperature of 1 kg of a substance through 1 K is known as specific heat capacity of the substance.
The unit of specific heat in this system is J ᐧ kg-1 ᐧ K-1
In SI, the specific heat capacity of water is 4186 J ᐧ kg-1 ᐧ K-1.
Relation between CGS and SI unit of specific heat capacity:
1 cal ᐧ g-1 ᐧ C-1 = \(\frac{4.186 \mathrm{~J}}{10^{-3} \mathrm{~kg} \times 1 \mathrm{~K}}\) = 4186 J ᐧ kg-1 ᐧ K-1
Specific heat capacities of a few solids and Liquids
Effect of the High Specific Heat of Water
In CGS system, the specific heat of water is 1 which is the highest among all solids and liquids. Therefore, 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. Hence, under identical conditions, water requires more time to be warmed up or cooled down in comparison with others. Due to this property. water is used as a cooling or a heating agent. As water needs a long time to be heated up, it is used in car radiators to keep the engine cool. Again as hot water needs a long time to cool down, it is used in heat reservoirs, in hot water bottles, etc.
Specific heat of sand, the main constituent of the earth’s surface, is very low. It warms up rapidly in daytime, while water in the sea remains comparatively cool. Thus the air over the land becomes hot and light and rises up, while the cool and heavier air over the sea flows towards the land setting up a sea breeze. At night, land cools faster than sea. The air over the sea surface being hot and light, rises up and the cool, heavier air over the land flows towards the sea setting up a land breeze.
Sea-shore and nearby areas remain comparatively cooler during the daytime and warmer during the night. Water in the sea takes a long time to be warmed up under the sun. Because of its high specific heat, water in the sea also takes a long time to cool at night. Hence, sea shore is neither very hot during the summer nor very cold during the winter.
Numerical Examples
Example 1.
Mass of an object is 200 g and its specific heat is 0.09 cal ᐧ g-1 °C-1. How much heat is required to increase its temperature from 20°C to 90°C?
Solution:
Here m = 200 g s = 0.09 cal ᐧ g-1 ᐧ C-1
t1 = 20°C t2 = 90°C
∴ Heat required,
H= ms(t2 – t1) = 200 × 0.09 × (90 – 20)
= 200 × 0.09 × 70 = 1260 cal.
Example 2.
In which case more heat is required?
(I) Temperature of 1 kg of water is raised from 30°C to 100°C.
(II) Temperature of 3 kg of iron, is raised from 30°C to 230°C. (specific heat of Iron = 0.12 cal g-1 ᐧ C-1).
Solution:
Heat gained in the first case
= m × s × (t2 – t1)
[m = 1000g, t1 = 30°C, t2 = 100°C]
= 1000 × 1 × (100 – 30)
= 1000 × 70 = 70000 cal
Heat gained in the second case
= m × s × (t2 – t1) [m = 3000 g, t1 = 30°C,
t2 = 230°C, s = 0.12 cal ᐧ g-1 ᐧ °C-1]
= 3000 × 0.12 × (230 – 30) = 72000 cal
Therefore in the second case, more heat is required.
Example 3.
It is given that the specific heat of water in cal ᐧ g-1 unit is s = 0.6t2, where t is the temperature in Celsius scale. How much amount of heat is required to raise the temperature of 1og water from 0°C to 10°C?
Solution:
Heat required to increase the temperature of water by an amount dt is,
dH = ms dt = m × 0.6t2dt
So, to raise the temperature of water from 0°C to 10°C, heat required
H = ∫ dH = \(\int_0^{10}\)m × 0.6t2 dt = m × 0.6\(\left[\frac{t^3}{3}\right]_0^{10}\)
= 10 × 0.6 × \(\frac{(10)^3}{3}\) = 2000 cal.
Effect of the High Specific Heat of Water
In CGS system, the specific heat of water is 1, which is the highest among all solids and liquids. Therefore, 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. Hence, under identical conditions, water requires more time to be warmed up or cooled down in comparison with others. Due to this property, water is used as a cooling or a heating agent. As water needs a long time to be heated up, it is used in car radiators to keep the engine cool. Again as hot water needs a long time to cool down, it is used in heat reservoirs, in hot water botdes, etc.
Specific heat of sand, the main constituent of the earth’s surface, is very low. It warms up rapidly in daytime, while water in the sea remains comparatively cool. Thus the air over the land becomes hot and light and rises up, while the cool and heavier air over the sea flows towards the land setting up a sea breeze. At night, land cools faster than sea. The air over the sea surface being hot and light, rises up and the cool, heavier air over the land flows towards the sea setting up a land breeze.
Sea-shore and nearby areas remain comparatively cooler during the daytime and warmer during the night. Water in the sea takes a long time to be warmed up under the sun. Because of its high specific heat, water in the sea also takes a long time to cool at night. Hence, sea shore is neither very hot during the summer nor very cold during the winter.
Numerical Examples
Example 1.
Mass of an object is 200 g and its specific heat is 0.09 cal ᐧ g-1. How much heat is required to increase its temperature from 20°C to 90°C?
Solution:
Here, m = 200 g, s = 0.09 cal ᐧ g-1 ᐧ C-1,
t1 = 20°C, t2 = 90°C
∴ Heat required,
H = ms(t2 – t1) = 200 × 0.09 × (90 – 20)
= 200 × 0.09 × 70 = 1260 cal.
Example 2.
In which case more heat is required?
(i) Temperature of 1 kg of water is raised from 30°C to 100°C.
(ii) Temperature of 3 kg of iron, is raised from 30°C to 230°C. (specific heat of iron = 0.12 cal ᐧ g-1 ᐧ °C-1).
Solution:
Heat gained in the first case
= m × s × (t2 – t1)
[m = 1000 g, t1 = 30°C, t2 = 100°C]
= 1000 × 1 × (100-30)
= 1000 × 70 = 70000 cal
Heat gained in the second case
= m × s × (t2 – t1) [m = 3000, t1 = 30°C, t2 = 230°C, s = 0.12 Cal ᐧ g-1 ᐧ C-1]
= 3000 × 0.12 × (230 – 30) = 72000 cal
Therefore in the second case, more heat is required.
Example 3.
It is given that the specific heat of water in cal ᐧ g-1 unit is s = 0.6t2, where t is the temperature in Celsius scale. How much amount of heat is required to raise the temperature of 10g water from 0°C to 10°C?
Solution:
Heat required to increase the temperature of water by an amount dt is,
dH = ms dt = m × 0.6t2dt
So, to raise the temperature of water from 0°C to 10°C, heat required,
H = ∫ dH = \(\int_0^{10}\) m × 0.6t2 dt = m × 0.6\(\left[\frac{t^3}{3}\right]_0^{10}\)
= 10 × 0.6 × \(\frac{(10)^3}{3}\) = 2000 cal.