WBBSE Solutions For Class 8 School Science Chapter 1 Physical Environment Heat Experiments

Chapter 1 Physical Environment Heat

Experiment -1

Let us take a beaker containing 100 grams of water. The beaker is placed over a wire gauge on a tripod stand. A thermometer is immersed into the beaker from a support, as shown in.

The initial temperature of the water is measured. Now the water is heated for some time till the temperature of the water is raised by 10°C.

If the water is heated for some more time, the increase in temperature is more. This means the more the time of heating, the more is heat energy supplied to the water and the more is the rise in temperature.

Inference: So, heat energy absorbed by the body is proportional to the rise in temperature of the body.

WBBSE Soulutions for class 8 Chapter 1 Physical environment thermometer

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Experiment-2

Let us now take two beakers one containing 100 grams of water and the other containing 200 gram of water. Both the beakers were separately heated for some time.

The experimental setup is the same as in the case of Experiment 1. The time of heating required to increase the temperature of water in both beakers by 10°C was noted.

It was found that the time required to raise the temperature of 200-gram water by 10°C is double the time required to increase the temperature of 100 grams of water by 10°C.

So, from the above experiment, it is found that the more the amount of water, the more the time of heating required to raise the temperature by a fixed amount.

Inference: So, heat energy absorbed by a body is proportional to the mass of the body.

WBBSE Soulutions for class 8 Chapter 1 Physical environment heat energy

Experiment-3

Let us now take two beakers – one containing 100 grams of water and the other containing 100 gram of oil. The two beakers were separately heated for the same amount of time. The experimental setup is the same as in the case of the Experiment

1. The increase in temperature in both beakers is noted. It is found that the rise in temperature in the case of water is less than that of oil.

So, from the above experiment, it is found that absorption of the same amount of heat energy by the same mass of different liquids produce different temperature change.

WBBSE Soulutions for class 8 Chapter 1 Physical environment Temeperature change

Inference: This means that heat energy absorbed by a body depends on the nature of the body (i.e. the chemical composition of the body).

On the basis of the above-mentioned experiments, we can propose a mathematical equation for the amount of heat energy absorbed by an object of mass “m”.

If the initial temperature of the object is “t” and after absorbing “Q” amount of heat, the temperature becomes “t” then we can write, Q = m.k. (t2 – tx)…………….1

Where “k” is a constant that depends on the nature (or material) of the object.

Unit of “Q” is calorie (in CGS unit) or joule (in SI unit). A cold body, when warmed, gains heat whereas a hot body, when cooled, loses heat. If t2 < t1, this implies a negative value of Q or heat lost by the object.

1. What are Calorie and Joule

The amount of heat required to raise the temperature of 1 gram of pure water by 1°C is 1 calorie.
Similarly, 1 joule is the amount of heat required to raise the temperature of 1 kilogram of pure water by 1 Kelvin (or 1°C).

So, when Q = 1 calorie; m = 1 gram;

(t2 -t1) = 1°C, then, 1 = 1 x k x 1

or, k = 1 calorie per gram per degree centigrade = 1 cal/g.°C

(For water)

2. Specific heat

“FC” is commonly called the specific heat and is generally expressed by “s”. The magnitude of V is different for different materials.

Mathematically we can say that when (t2 – t1) = 1°C and “m” is equal to 1, then “s” is equal to “Q”. This gives us the definition of specific heat.

Definition of specific Heat: Specific heat of a substance is the heat required to raise the temperature of the unit mass of the substance by 1°C.

So, a unit of “s” is calorie per gram per degree centigrade (cal/g.°C) (in CGS unit) or joule per kilogram per degree centigrade (J/kg.°C) (in SI unit). We can now rewrite equation 1. as, Q = m.s. (t2-t1)

2. That is, Heat energy absorbed (or lost) by a body = mass of the body x specific heat of the body x change in temperature.

Substance Specific heat (cal/g.°C)
Aluminium 0.215
Carbon 0.121
Copper 0.0923
Silver 0.0564
Lead 0.0305

Water has the highest specific heat among the Commonly known substances (except a few).

3. Law of mixtures

Generally, when a hot substance is mixed with a cold one, an exchange of heat takes place. The body at a higher temperature loses heat to the body at a lower temperature.

If no heat is lost to the surroundings, then, Heat gained = Heat lost This relation follows from the law of conservation of energy.

 

WBBSE Soulutions for class 8 Chapter 1 Physical environment state of subastance

 

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