Heat Transfer

HEAT TRANSFER

Heat transfer is the process of transfer of heat from high temperature reservoir to low temperature reservoir. In terms of the thermodynamic system, heat transfer is the movement of heat across the boundary of the system due to temperature difference between the system and the surroundings

TEMPERATURE

Temperature is the measure of hotness and coldness

CELSIUS TO FAHRENHEIT

(C × 9/5) + 32 = °F

FAHRENHEIT TO CELSIUS

(F- 32)5/9 = °C

CELSIUS TO KELVIN

C°+273 = K

THREE TYPES OF HEAT TRANSFER

Heat is transferred via ------

* solid material (conduction)

ex. 

touching a stove and being burned

* liquids and gases (convection)

ex.

Hot air rising, cooling, and falling

* electromagnetic waves (radiation)

ex. sun, the heat coming from the fire from far away

Key Terms
condensation -the conversion of a vapor or gas to a liquid. 

freezing -turning into very cold temperature

melting-the conversion of solid liquid

plasma -is a state of matter in which an ionized gaseous substance becomes highly electrically conductive to the point that long-range electric and magnetic fields dominate the behaviour of the matter. The plasma state can be contrasted with the other states: solid, liquid, and gas

sublimation-  is the transition of a substance directly from the solid to the gas phase, without passing through the intermediate liquid phase.

vaporization -a phase transition from the liquid phase to vapor. There are two types of vaporization: evaporation and boiling. Evaporation is a surface phenomenon, whereas boiling is a bulk phenomenon.

SPECIFIC HEAT AND LATENT HEAT

phase change -a change from one state (solid or liquid or gas) to another without a change in chemical composition. phase transition, physical change, state change. freeze, freezing - the withdrawal of heat to change something from a liquid to a solid. liquefaction - the conversion of a solid or a gas into a liquid.


SPECIFIC HEAT  

the heat required to raise the temperature of the unit mass of a given substance by a given amount (usually one degree).

SPECIFIC HEAT FORMULA IS

m C ΔT

m = mass
c = specific heat constant
ΔT = (T2 - T1) as temperature

Question: A 500 gram cube of lead is heated from 25 °C to 75 °C. How much energy was required to heat the lead? The specific heat of lead is 0.129 J/g°C.

Solution: First, let’s the variables we know.

m = 500 grams
c = 0.129 J/g°C
ΔT = (Tfinal – Tinitial) = (75 °C – 25 °C) = 50 °C

Plug these values into the specific heat equation from above.

Q = m c ΔT

Q = (500 grams)·(0.129 J/g°C)·(50 °C)

Q = 3225 J

Answer: It took 3225 Joules of energy to heat the lead cube from 25 °C to 75 °C.

Question: A 25-gram metal ball is heated 200 °C with 2330 Joules of energy. What is the specific heat of the metal?

Solution: List the information we know.

m = 25 grams
ΔT = 200 °C
Q = 2330 J

Place these into the specific heat equation.

Q = mcΔT

2330 J = (25 g)c(200 °C)

2330 J = (5000 g°C)c

Divide both sides by 5000 g°C

c = 0.466 J/g°C

Answer: The specific heat of the metal is 0.466 J/g°C.

LATENT HEAT

latent heat of fusion - The amount of heat energy released or absorbed when a solid changing to liquid at atmospheric pressure at its melting point is known as the latent heat of fusion. Ex: The latent heat of fusion of water is 79.72 cal/gram (or) 334.0 kJ/Kg.

latent heat of vaporization - is a physical property of a substance. It is defined as the heat required to change one mole of liquid at its boiling point under standard atmospheric pressure. It is expressed as kg/mol or kJ/kg. ... The heat of vaporization of water is about 2,260 kJ/kg, which is equal to 40.8 kJ/mol

m Lf   or  m Lv

m = mass                                                                                

Lf  = latent heat of fusion
Lv = latent heat of vaporization

Sample Problem

Calculate the amount of heat added to 1 gram gold to change phase from solid to liquid. The heat of fusion for gold is 64.5 x 10³ J/kg.

Known :

Mass (m) = 1 gram = 1 x 10^-3 kg

Heat of fusion (LF) = 64.5 x 10³ J/kg

Wanted : Heat (Q)

Solution :

Q = m LF

Q = (1 x 10^-3 kg)(64.5 x 10³ J/kg)

Q = 64.5 Joule

 Calculate the amount of heat released by 1 gram mercury to change phase from liquid to solid. Heat of fusion for mercury is 11.8 x 10³ J/kg.

Known :

Mass (m) = 1 gram = 1 x 10^-3 kg

Heat of fusion (LF) = 11.8 x 10³ J/kg

Wanted : Heat (Q)

Solution :

Q = m LF

Q = (1 x 10^-3 kg)(11.8 x 10³ J/kg)

Q = 11.8 Joule

Gas

Gas is a substance or matter in a state in which it will expand freely to fill the whole of a container, having no fixed shape (unlike a solid) and no fixed volume (unlike a liquid

Gas Law

The gas laws were developed at the end of the 18th century, when scientists began to realize that relationships between pressure, volume and temperature of a sample of gas could be obtained which would hold to approximation for all gases.

The Gas Laws: Pressure Volume

Temperature Relationships

Boyle's Law:  The Pressure-Volume Law

Boyle's law or the pressure-volume law states that the volume of a given amount of gas held at constant temperature varies inversely with the applied pressure when the temperature and mass are constant.

Another way to describing it is saying that their products are constant.

PV = C

When pressure goes up, volume goes down. When volume goes up, pressure goes down. From the equation above, this can be derived:

P₁V₁ = P₂V₂ = P₃V₃ etc.

This equation states that the product of the initial volume and pressure is equal to the product of the volume and pressure after a change in one of them under constant temperature.  For example, if the initial volume was 500 mL at a pressure of 760 torr, when the volume is compressed to 450 mL, what is the pressure? Plug in the values: P₁V₁ = P₂V₂ (760 torr)(500 mL) = P₂(450 mL) 760 torr x 500 mL/450 mL = P₂ 844 torr = P₂ The pressure is 844 torr after compression.

Charles' Law:  The Temperature-Volume Law

This law states that the volume of a given amount of gas held at constant pressure is directly proportional to the Kelvin temperature. V T Same as before, a constant can be put in: V / T = C As the volume goes up, the temperature also goes up, and vice-versa. Also same as before, initial and final volumes and temperatures under constant pressure can be calculated. V₁ / T₁ = V₂ / T₂ = V₃ / T₃ etc.

Gay-Lussac's Law:  The Pressure Temperature Law

This law states that the pressure of a given amount of gas held at constant volume is directly proportional to the Kelvin temperature. P T Same as before, a constant can be put in: P / T = C As the pressure goes up, the temperature also goes up, and vice-versa. Also same as before, initial and final volumes and temperatures under constant pressure can be calculated. P₁ / T₁ = P₂ / T₂ = P₃ / T₃ etc.

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