Heat

TOPIC 4 :HEAT4.2 : Specific Heat Capacity4.3 : Specific Latent Heat4.4 : Gas Laws4.1 : Thermal Equilibrium

HOT!HOT!HOT!COLD!!!!!!!!

Take a cube of ice. Ask students to hold the ice until its melt.Then, ask students why the ice was melt.

Object AObject BHeat energy is transferred between two objectEnergy transferred at a faster rate from hot object A to the cold object BA and B in thermal equilibrium.

There is no net of heat between two object that are in thermal equilibrium. Two objects in thermal equilibrium have the same temperature.

Two objects at different temperatures in thermal contact will eventually come to a state of thermal equilibrium. It does not depend on theMass* sizeType of material* shape

Thermal EquilibriumThermometerCelsius scaleFixed pointLiquid-in-glassthermometerResistance thermometerThermocoupleThermometer scaleUpper fixed pointLower fixed pointFundamental intervalL100 = 100CL0 = 0CSteam pointIce point1C = L100 L0 100

The liquid used in liquid in glass thermometersshould :-a). Be easy seenb). Expand and contract rapidly over a wide range of temperaturec). Not stick to the glass wall of the capillary tube

Mercury in a glass,Will expand if heated,Flow in capillary tube,Up and down with temperature, Scale shows tenth, Glass bulb thin-walled, Mercury heats up quickly, Clinical thermometer thats it.Clinical Thermometer

Thermometer scaleFixed point The lower fixed point is ice point = 0C The upper fixed point is steam point = 100C Distance in betweens calls fundamental interval.

Lower fixed pointL0Upper fixed pointL100Fundamental interval1C = L100 L0 100

Experiment Set Up :Fundamental intervalLower fixed pointL0Upper fixed pointL0SteamBoiling waterwaterIce meltingconeRound-bottomed flaskBeakerTHERMOMETER CALIBRATING(a) (b)

Apparatus Water ice rubber stopper Thermometer without scale Wire gauzeRound-bottomed flaskBeakerCone Bunsen burner

Procedure :Ice-pointa) Put in thermometer without scale into the melting ice as shown in the figureb) Determine the level of mercury in the thermometer without scalec) Measure the length of mercury column, L0

2. Steam pointArrange the apparatus shown has in the figureb) Fixed the highest level at the thermometer without scale when the water is boilingc) Measure the length of the mercury level, L100

3. Divide the scale between ice-point and steam point to 100 division at equal intervals. Each division equals to 1C

Results:Given, L0 = 3.5 cmL100 = 15.6 cmThe column of mercury= 12.2 cm= 71.9C = 71.9 C

Discussion :

Expansion of the volume of mercury when the temperature increases

Conclusion :When calibrate the thermometer, two standard points that is heated for ice-point and steam point are chosen. The range of the two levels that is divided into divisions is used to measure the temperature between these levels

Questions :What is temperature?Name the types of thermometer?How to calibrate the liquid-in-glass thermometer?How to determine the distance between lower and upper fixed points?The distance between ice point and steam point is 25cm, determine the temperature as shown in the figure below.0C100C12 cm

candleBurn the wooden spoon and steel spoon on a lighted candle for about 3 minutes. Specific Heat Capacity

What do you feel ?

I can feel that steel spoon is hotter than wooden spoon.

Why ?

Because the steel spoon have a greater Heat Capacity than wooden spoon.

The quantity of heat absorbed is depend on the type of object or the type of substance. For example steel has smaller heat capacity than wood.

EXAMPLEPan is made from high steel which has lower heat capacity.Handles of cooking utensils are made from plastic which has high heat capacity.

Handle of kettle is made from high heat capacity material to enable consumer to hold or lift during its hot. EXAMPLE

A women in the kitchen finds that when the same amount of water and cooking oil are heated separately, the cooking oil become hot faster than the water.

The water and oil are said to have different specific heat capacity.

What is the Specific Heat Capacity?The heat energy take in or given out when 1 kg of a substance changes temperature by 1C Q = Heat energy m = mass = initial and final temperature respectivelyS.I unit -- J kg C

ObjectSpecific Heat CapacityThe rate of temperature increase ALowFast

BHighSlow

EXPERIMENT :Aim : To find the specific heat capacity of water.Objective : To determine the specific heat capacity of water.

Apparatus :12V Power supplyHeaterThermometerStop watchAsbestosHeat insulationStirrerBeakerRod

Figure :WaterHeat insulationAsbestosThermometerStirrerHeater

Method :Set up the apparatus as shown in figure.Weigh accurately an empty beaker and determine the mass (m1).Place the water into the beaker and weigh accurately again. Record the mass of beaker and water (m2).Take the initial temperature of water (1). Switch on the power supply.Stir the water simultaneously with the stirrer.Record the final temperature (2) after t minutes.

Specific heat = 60Ptcapacity, c(m2m1)(1 2)Jkg-1c-1

Calculation :Power of heater : P WattMass of water : (m2 - m1)gDuration of heating : t x 60 sDifference of temperature, = 1 2

The value of C that you get is different from the actual of Specific Heat Capacity, It is because some of the heat is lost to the surrounding.

Discussion

The value of Specific Heat Capacity is 4200 J kg -1C-1

Conclusion :

Activity : Observe the change of temperature

To observe the change in temperature when the same amount of heat is used to heat different masses of waterThe water with the mass 0.5 kg has a _____ heat capacity than the water with the mass 2.0 kg.

The larger the mass of the water, _____ is its heat capacity

Specific Heat Capacity Q = mc Ep = mgh Ek = 1/2mv E = Ptmgh = mc 1/2mv = mcPt = mc E = Qm =C==

ExampleHow much heat energy is required to raise the temperature of 5 kg iron bar from 32 C to 62 C? ( specific heat capacity of iron is 452 J kg C ).

2. A bottle containing 1.5 kg of water at 35 C is put into a refrigerator. What is the temperature of the water after 157500 J of heat has been removed from the water ? ( C water = 4200 J kg C )

3. A block of copper with a mass of 5 kg is cooled from 100 C to room temperature 27 C. How much heat is released by the copper block ? ( C of copper is 380 J kg c )

4. Determine the heat capacity of a liquid if 300 kJ is required to heat 2 kg of the liquid from 25 C to 85 C

5. A 100 W electric heater is used to heat 4 kg of olive oil for 10 minutes. Assuming that there is no loss of heat, calculate the temperature rise of the olive oil. The specific heat capacity of olive oil is 1890 J kg C.

6 .The water temperature at the top of a 200 m high water is 20 C. What is the water temperature at the bottom of the waterfall ? ( g= 10 ) ( c water = 4200)

7. A bullet traveling at 200 m s hit a sand bag. The temperature of the bullet rises by 50 C. Assuming hat the kinetic energy of the bullet is converted into heat energy which up the bullet, calculate the specific heat of the bullet.

Specific Latent Heat

candleice

Specific Latent Heatthe amount of heat required to change the phase of 1 kg of substance at a constant temperature.l = Q/ mQ = heat / energym = massl = Specific Latent HeatSI unit - J kg

GasLiquidLatent heat taken inLatent heat given outSolid

Particles vibrate about fix positionsParticles vibrate and move each otherParticles vibrate about fix positionsLatent Heat of FusionDuring melting, the temperature remains constant although heat continues to be supplied because the heat absorbed is used to break up the bonds between the particles

Latent Heat of VaporizationParticles vibrate about fix positions

Particles vibrate & move among each other

During boiling, the temperature remains constant although heat continues to be supplied because the heat absorbed is used to break up the bonds between the particles

Example : The heating curve for waterLatent heat of fusionLatent heat of vaporization

Electronic balanceAfter 5 minutesExperiment figure : Specific Latent Heat of vaporization

E= PtEp = mgh Ek = mvE = QQ = mlPt = ml

ExamplesAn electric kettle contains 3.4 kg of water. calculate the amount of heat requite to boil away all the water after boiling point has been reached.If the power of the heater is 2.4 kW, what is the time taken?( specific latent heat of vaporization of water is 2260000 J kg )

Questions A solid X with a mass of 200 g is heater at uniform rate by an electric heater of 150 W. The temperature-time graph for this heating process is shown in figure.2010080604024861012ABCD

Which point on the graph shows the beginning of the melting process?

(b) What is the melting point of X?

Define specific heat capacity and latent heat of fusion.

(d) By utilizing the graph in figure, calculate(i) the specific heat capacity of X(ii) the latent heat of fusion of X

Explain why the temperature of X is constant even though heat is continuously supplied to it.

Specific Latent HeatConclusion :

the transfer of heat during a change of phase does not cause a change in temperature. specific latent heat of vaporization of water = 2.26 X 106 Jkg-1 specific latent heat of fusion of water = 3.36 X 105 Jkg-1

Gas lawsPressure LawCharless LawBoyles LawT , constantV , constantP , constant

Boyles LawWhat happen to the balloon when we press it like shown in the photo?

Boyles LawTake one syringe. Put the syringe between your two palm. And press the piston.

Observe the change of volume whenthey change the pressure.

Figure :Rubber tubeSyringeBourdon GaugeGas

Experiment Figure :clampsyringerubber tubeclamp

Boyles Law said :the volume of a fixed mass of gas at constant temperature is inversely proportional to the pressureP1 V1 = P2 V2P = pressureV = volumeT = temperature

Charless Law said :the volume of a fixed mass of gas at constant temperature is directly proportional to the pressureV1 = V2T1 T2

Pressure Law said :the pressure of a fixed mass of gas at constant volumeis directly proportional to the TemperatureP1 = P2T1 T2

Questions :What are the type of gas laws?What is the Boyles Law?What is the Charless Law?What is the Pressure Law?A syringe containing 5ml of air at room temperature, 27C. Then the syringe was kept in a refrigerator and the temperature decreased to 3C. What was the volume of air in the syringe?T1= 27CT2= 3CV1=5mlV2