Calorimetry

Calorimetry

We studied about the effects of heat energy. When heat energy is supplied to substances, physical changes take place in them. Solid form of water (ice) is changed to liquid form, and liquid form of water is changed to gaseous form. These are all the physical changes due to heat energy. Similarly, heat energy produces chemical changes also. To know more about the physical and chemical changes that take place in substances, we need to measure the amount of heat involved. The technique used to measure the amount of heat involved in a physical or a chemical process is known as calorimetry.

1. Temperature

Temperature is a physical quantity which expresses whether an object is hot or cold. It is measured with the help of thermometer. There are three scales to measure the temperature. They are:

* Celcius scale

* Fahrenheit scale

* Kelvin scale

Among these three scales, Kelvin scale is the most commonly used one. You will study about this elaborately in Standard IX.

2. Unit of Heat

We know that heat is a form of energy. The unit of energy in SI system is joule. So, heat is also measured in joule. It is expressed by the symbol J. The most commonly used unit of heat is calorie. One calorie is the amount of heat energy required to raise the temperature of 1 gram of water through 1°C. The relation between calorie and joule is given as, 1 calorie = 4. 186 J.

The amount of energy in food items is measured by the unit kilo calorie.

 1 kilo calorie = 4200 J (Approximately).

3. Heat capacity

Activity 6

Take some amount of water and cooking oil in two separate vessels. Heat them till they reach a particular temperature (Caution: Heat the oil under the supervision of your teacher). Which one is heated first? Water will take more time to get heated. Why?

Answer:

(i) Heat transfer depends on the nature of the substance.

(ii) Water has high specific heat capacity than that of cooking oil.

(iii) A substance with high specific heat capacity absorbs a large quantity of heat.

(iv) Thus, it takes long time to heat up.

In general, the amount of heat energy gained or lost by a substance is determined by three factors. They are:

* Mass of the substance

* Change in temperature of the substance

* Nature of the material of the substance

Different substances require different amount of heat energy to reach a particular temperature. This nature is known as heat capacity of a substance. Heat capacity is defined as the amount of heat energy required by a substance to raise its temperature by 1°C or 1 K. It is denoted by the symbol C'.

Heat capacity = Amount of heat energy required (Q) / Raise in temperature (ΔT)

 C'  = Q / ΔT

 The unit of heat capacity is cal / °C. In SI system, it is measured in JK^-1.

Water has higher heat capacitythan most other substances.This accounts for the use of water as common coolant.100 g of water can take away more heat than 100 g of oil.

Problem 1

The temperature of a metal ball is 30° C. When an energy of 3000 J is supplied, its temperature raises by 40° C. Calculate its heat capacity.

Solution

Heat capacity, C' = Q / ΔT

Here, Q = 3000 J

ΔT = 40°C - 30°C = 10°C or 10 K

C' = 3000 / 10 = 300 JK^-1

The heat capacity of the metal ball is 300 JK-1.

Problem 2

The energy required to raise the temperature of an iron ball by 1 K is 500 JK^-1. Calculate the amount of energy required to raise its temperature by 20 K.

Solution

Heat capacity, C' = Q / ΔT

 Q=C'×ΔT

Here, C' = 500 JK^-1

 ΔT=20K

 Q = 500 × 20 = 10000 J.

The amount of heat energy required is 10000 J.

4. Specific heat capacity

When the heat capacity of a substance is expressed for unit mass, it is called specific heat capacity. Specific heat capacity of a substance is defined as the amount of heat energy required to raise the temperature of 1 kilogram of a substance by 1°C or 1 K. It is denoted by the symbol C.

Specific heat capacity  = { Amount of heat energy required (Q) } / { Mass × Raise in temperature (ΔT) }

C = Q / m × ΔT

The SI unit of specific heat capacity is J kg^-1 K ^-1.

Problem 3

An energy of 84000 J is required to raise the temperature of 2 kg of water from 60° C to 70° C. Calculate the specific heat capacity of water.

Solution

Specific heat capacity, C = Q / m × ΔT

Here, Q = 84000 J

 m = 2 kg

 ΔT = 70° C – 60° C = 10° C or 10 K

C = 84000 / 2×10 = 4200 J kg^-1 K^-1

The Specific heat capacity of water is 4200 J kg^-1 K^-1.

Problem 4

The specific heat capacity of a metal is J kg^-1K^-1. Calculate the amount of heat energy required to raise the temperature of 500 gram of the metal from 125° C to 325° C.

Solution

Specific heat capacity, C = Q / m × ΔT

 Q = C × m × ΔT

Here, C = 160 J kg K^-1

m = 500 g = 0.5 kg

ΔT = 325° C – 125° C = 200° C or 200 K

 = 160 × 0.5 × 200 = 16000 J.

The amount of heat energy required is 16000 J.