HBSE Notes for Heats of Reactions and Calorimetry

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Calorimetry 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.

Numerical Problems On Heats Of Reactions Class 11 HBSE

  • 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.

Calorimetry Effect Of The High Specific Heat Of Water Numerical Examples

Example 1. Mass of an object is 200 g and its specific heat is 0. 09 cal · g-1 · °C-1. How much heat is required to increase its temperature from 20°C to 90°C?
Solution:

Given

Mass of an object is 200 g and its specific heat is 0. 09 cal · g-1 · °C-1.

Here, m = 200 g, s = 0.09 cal · g-1 · °C-1,

t1 = 20°C, t2 = 90°C

∴ Heat required,

H = ms(t2-t1) = 200 x 0.09 x (90-20)

= 200 x 0.09 x 70 = 1260 cal.

1260 cal heat is required to increase its temperature from 20°C to 90°C

Heats Of Reactions And Calorimetry Class 11 HBSE Notes

Example 2. In which case more heat is required?

  1. The temperature of 1 kg of water is raised from 30° C to 100°C.
  2. Temperature of 3 kg of iron, is raised from 30°C to 230°C. (specific heat of iron = 0.12 cal · g-1 · °C-1).

Solution:

Heat gained in the first case = m x s x (t2 – t1)

[m = 1000 g, t1 = 30°C, t2 = 100°C]

= 1000 x 1 x (100-30)

= 1000 x 70 = 70000 cal

Heat gained in the second case

= m x s x(t2– t1) [m = 3000 g, t1 = 30°C, t2 = 230°C, s = 0.12 cal · g-1 · °C-1]

= 3000 x 0.12 x (230 – 30) = 72000 cal

Therefore in the second case, more heat is required.

Definition Of Heats Of Reactions And Calorimetry Class 11

Example 3. It is given that the specific heat of water in cal · 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:

Given

It is given that the specific heat of water in cal · g-1 unit is s = 0.6t², where t is the temperature in Celsius scale.

Heat required to increase the temperature of water by an amount dt is,

dH = ms dt = mx 0.6t² dt

So, to raise the temperature of water from 0°C to 10°C, heat is required,

H = \(\int d H=\int_0^{10} m \times 0.6 t^2 d t=m \times 0.6\left[\frac{t^3}{3}\right]_0^{10}\)

= \(10 \times 0.6 \times \frac{(10)^3}{3}=2000 \mathrm{cal} .\)

Heat Of Reaction Formula In Calorimetry Class 11 HBSE

Calorimetry Thermal Capacity Or Heat Capacity

Thermal Capacity Of A Body Or Heat Capacity Of A Body Definition: Thermal capacity of a body is defined as the quantity of heat required to raise its temperature by unity.

Let the mass of a body be m and its specific heat be 5. Heat gained by the body to raise its temperature by unity is H = m s · 1 = ms. By definition, this is the thermal capacity. It is usually denoted by the symbol C.

Hence, thermal capacity of a body = mass of the body x specific heat capacity of its material.

Units of thermal capacity:

  1. cal · °C-1 CGS System
  2. J · K-1

Importance Of Calorimetry In Heat Reactions Class 11

Relation between thermal capacity and specific heat: We know, thermal capacity of a body

= mass of the body x specific heat of its material

or, specific heat of the material

= \(\frac{\text { thermal capacity of the body }}{\text { mass of the body }}\)

If the mass of the body is 1 unit, the specific heat of the material = thermal capacity of the body.

Hence, thermal capacity of the unit mass of a body is the specific heat of its material.

 

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