HEAT

HEAT

1. Fill in the blanks and rewrite the sentence.

a)  The amount of water vapour in air is determine in terms of its_______________ (absolute humidity)

b)  If objects of equivalent masses are given equal heat, their final temperature will be different.      This is due to in their___________________ (specific heat capacity)

c) During transformation of liquid phase to solid phase the latent heat is______________ (given out)

Latent heat

Latent heat can be understood as heat energy in hidden form which is supplied or extracted to change the state of a substance without changing its temperature.

                                           

Normally when heat energy is added to or remove from an object the temperature of the object change; however during phase change the temperature of an object stays constant. The temperature remain the same because energy is required for an object to change phases.

Q2) Observe the following graph considering the change in volume of water as its temperature is raised from 0°C, discuss the difference in the behaviour of water and other Substances. What is this behaviour of water called?

                    

Ans:- 1) The above graph tell us that water is heated from 0°C its volume goes on decreasing up to 4°C. It is minimum at 4°C if it is further heated its volume goes on increasing.

2) If we take the case of any other substance, it expand on heating. But water contract up to 4°C than above 4°C it start expanding.

3) The behaviour of water between its temperatures from 0°C to 4°C is called anomalous behaviour of water.

Regelation

The phenomenon due to which ice melts to water at a temperature below 0°C melts to water on application of external pressure and refreezing of water formed into ice on the released of pressure is called regelation

Examples of Regelation :

As we may have seen the preparation of an ice-ball. First, an ice slab is shredded and then the shredded ice is pressurised around the tip of a stick to prepare the ice-ball.

If two small pieces of ice are taken and Pressed against each other for a while, they stick to each other.

Define the terms 1) latent heat of fusion 2) latent heat of vaporization

Latent heat of fusion: - The amount of heat energy absorbed at constant temperature by unit mass of a solid to convert into liquid phase is called the specific latent heat of fusion.

Latent heat of vaporization: - The amount of heat energy absorbed at constant temperature by unit mass of a liquid to convert into gaseous phase is called the specific latent heat of vaporization.

Anomalous behaviour of water

In general, when a liquid is heated up to a certain temperature, it expands, and when Cooled it contracts. Water, however, shows a special and exceptional behaviour. If we heat water from 0°C up to 4°C, it contracts instead of expanding. At 4°C its volume is minimum (due to contraction). If the water is heated further, it expands and its volume increases. The behaviour of water between its temperatures from 0°C to 4°C is called anomalous behaviour of water.

Hope’s Apparatus

The anomalous behaviour of water can be studied with Hope’s apparatus. In Hope’s apparatus.

Construction:-

A flat bowl is attached to a cylindrical container as shown in figure. There is provision to attach two thermometers above (to measure temperature T2) and below (to measure temperature T1) the flat bowl on the cylindrical container. Water is filled in the cylindrical container and a mixture of ice and salt (freezing mixture) is put in the flat bowl.

 

Hope’s Apparatus

Observation:-

•        During the study of anomalous behaviour of water using Hope’s apparatus, temperature T1 and T2 are recorded after every 30 seconds.
•        The temperatures are plotted on the Y-axis and the time in minutes on the X-axis. The graph is shown in figure. The graph shows that initially, both the temperatures T1 and T2    are identical.

Time- temperature graph

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•        However, as time passes, temperature T1 of water in the lower part of the cylinder decreases fast, while, temperature T2 of water in the upper part of the cylinder decreases comparatively slowly 
•       Once the temperature T1 of the lower part reaches 4°C it remains almost stable at that temperature.
•       T2 decreases slowly to 4°C thereafter, since T2 starts changing rapidly it records 0°C first and after that the lower thermometer T1 records 0°C temperature.
•      The point of intersection of the two curves shows the temperature of maximum density.

Explanation:-

•            Initially, the temperature of water in the middle of cylinder lowers due to freezing mixture in the outer bowl.
•            Since the temperature of water there decreases, the water contract so its volume decreases, and its density increases.
•         The water with higher density moves downwards. Therefore, the lower thermometer T1 shows rapid fall in temperature and this continues till the temperature of water becomes equal to 4°C.
•         When the temperature of water starts decreasing below 4°C, its volume increases, and density decreases.  Therefore it moves in the upward direction.
•        The temperature of water in upper part (T2), therefore decreases rapidly to 0°C. The temperature of water in the lower part (T1), however, remains at 4°C for some time and then decreases slowly to 0°C.

       Dew point and Humidity

Humidity: - The presence of water vapor in the air makes it moist. The moisture in the atmosphere is called humidity.

For a given volume of air, at a specific temperature, there is a limit on how much water vapour the air can contain. If the amount exceeds this limit, the excess vapor converts into water droplets. When the air contains maximum possible water vapor, it is said to be saturated with vapour at that temperature.

For Example:-

If temperature of air is 40°C, it can contain 49 grams of water vapour per kilogram of dry air without condensation. If the amount of vapour exceeds this limit, the additional vapour will condense. However, if the temperature of air is 20°C, it can contain only 14.7 grams of water vapour per kilogram of dry air without condensing. If the vapour contained in air is less that the maximum limit, then the air is said to be unsaturated.

Unit of heat

The units of heat are Joule (J) in SI units, cal (calorie) in CGS units.  

  Ø  The amount of heat necessary to raise temperature of 1 g of water by 1°C from 14.5°C to 15.5°C is called one cal heat.

Similarly,

  Ø  The amount of heat necessary to raise the temperature of 1 kg of water by 1°C from 14.5°C to 15.5°C is called one kcal heat.

  Ø It is clear that (1 kcal= 1000 cal).

      Specific Heat Capacity

      The amount of heat energy required to raise the temperature of a unit mass of an  object by 1°C is called the specific heat of that object. 

  Ø The specific heat capacity is denoted by letter ‘c’. The SI unit of specific heat is J/ °C kg, and the CGS unit is cal/g °C

     If specific heat of an object is ‘c’, the mass of the object is ‘m’ and if the temperature of the object is raised by ∆T °C, the heat energy absorbed by the object or heat energy loss by the object is given by 
        ( m × c × ∆T )

       Heat Exchange

       Heat energy lost by the hot object = Heat energy gained by the cold object

       The specific heat of an object can be measured using mixing method. For this calorimeter is used. If a hot solid object is put in the water in a calorimeter, heat exchange between the hot object and the water and calorimeter starts. This continues till the temperatures of the solid object, water and the calorimeter become equal. Therefore,

     Heat lost by solid object = heat gained by water in calorimeter + heat gained by the calorimeter.

     Here, heat lost by the solid object (Q) = mass of the solid object × its specific heat × decrease in its temperature.

     Similarly,

     Heat gained by the water (Q1) = mass of the water × its specific heat × increase in its temperature

      Heat gained by the calorimeter (Q2) = mass of the calorimeter × its specific heat × increase in its temperature.

     Heat lost by hot object = Heat gained by calorimeter + Heat gained by water.

Q = Q2+ Q1

      Using these equations, if the specific heat of water and the calorimeter are known, the specific heat of the solid object can be calculated.

   

    Example 1: How much heat energy is necessary to raise the temperature of 5 kg of water from 20°C to 100°C?

     Given: m= 5 kg, c = 1 kcal/kg°C and change in temperature ∆T = 100-20 = 80°C

     Energy to be supplied to water = energy gained by water

     = mass of water × specific heat of water × change in temperature of water

     = m × c × ∆T

      = 5 × 80°C

      = 400 kcal 

      Hence, the heat energy necessary to raise the temperature of water = 400 kcal.

   

      Example 2: A copper sphere of 100 g mass is heated to raise its temperature to 100°C and is released in water of mass 195 g and temperature 20°C in a copper calorimeter. If the mass of calorimeter is 50 g, what will be the maximum temperature of water?

      Given: Specific heat of copper = 0.1 cal/g°C

                  And so specific heat of calorimeter= 0.1 cal/ g°C

     Suppose the copper ball water and the calorimeter attain final temperature T.

      

     Heat lost by solid object = heat gained by water in calorimeter + heat gained by the calorimeter.

      Here, heat lost by the copper ball = mass of the copper × specific heat of copper × decrease in temperature of the ball

       Q = 100 × 0.1 × (100 - T)

      Similarly,

      Heat gained by the water = mass of the water × its specific heat × increase in its temperature

      Q1 = 195 × 1 × (T - 20) and

      Heat gained by the calorimeter = mass of the calorimeter × its specific heat × increase in its temperature

      Q2= 50 × 0.1 × (T - 20)

      Q = Q1 + Q2

      100 × 0.1 × (100 - T) = 195 × 1 × (T - 20) + 50 × 0.1 × (T - 20)

      10 (100 - T) = 195 (T - 20) + 5 (T - 20)

      10 (100 - T) = 200 (T - 20)

2    10 T = 5000

      T = 23.8°C

      ∴ the maximum temperature of water will be 23.8°C.

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