HEAT AND THERMAL ENERGY: Rate of Heat Transfer Within Various Materials

Thus far, discussions about the rate at which heat is transferred to a system has been ignored. We know from experience that materials that are removed from a hot oven will retain heat energy differentially in accordance with their subatomic structure. Fortunately, such observations have been used to construct a very modest yet useful expression that relates the rate of heat transfer ( H ) to several observable quantities:

H = kAΔT / L

The ( A ) variable represents the cross-sectional area of an object through which heat transfer occurs, ΔT represents the temperature difference between the object and source of heat, and ( L ) is in reference to the length of the object at hand. The ( k ) variable represents the thermal conductivity constant, and its value differs from one material to another. Heat energy transferred into a medium over a certain period of time ( Δt ) is quantified with the SI unit of power, which is the watt ( W ):

Power ( W ) = ( joules / second ) = Q / Δt

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Q: A young child walks across a surface made of stones onto a surface made of wood. Upon doing so, she immediately notices that the wooden surface feels “ warmer “. What may be the cause of this?

A: Stone has a higher conductivity constant than wood; thus, heat leaves the child’s foot at a greater rate when it is in contact with the stone surface. The molecular arrangement of atoms and molecules in wood has less power to remove heat at the same rate.

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Q: A 20 cm-thick glass window of a furnace is exposed to a temperature of 1,100^o C on one side, while a temperature of 200^o C is maintained on the other side. What is the heat transfer rate per unit area through the window?

A: The thermal conductivity constant ( k ) value of the glass window is ( 0.9 J / s*m*K ).

Note: The ( m ) variable stands for meters, not mass!!! When ( A ) is divided by ( L ), we are left with the unit for length in the numerator of the expression.

H = kAΔT / L

H / A = kΔT / L

kΔT / L = [ ( 0.9 J / s*m*K )( 1,100^o C – 200^o C ) ] / ( 0.2 m )

kΔT / L = [ ( 0.9 J / s*m*K )( 900^o C ) ] / ( 0.2 m )

Note: Fortunately, change in temperature on the Celcius scale is directly proportional to changes in temperature on the Kelvin scale, so we may substitute terms accordingly.

kΔT / L = [ ( 0.9 J / s*m*K )( 900^o K ) ] / ( 0.2 m )

kΔT / L = 4,050 W / m²

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Q: How are watts-per-square-meter units used in studies of sound and light?

A: Intensity ( I ) = Watts per unit area

I = W / m²

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Q: A 1.5 meter stainless steel rod is exposed to a fire that has a temperature of 850 K. The rod has a cross-sectional radius of 1 cm, and the cooler end of the rod has a temperature of 300 K. What is the rate of heat transfer through the rod?

Note: It may be useful ( albeit inaccurate ) to compare the temperature differential to the potential difference that exists between charged plates of a capacity ( or any electrical potential difference, for that matter ). If a great potential difference exists, it will give rise to proportionally greater current flow than systems where the voltage is of a lower magnitude.

A: A = πr²

A = ( 3.14 )( 0.01 m )²

A = ( 3.14 )( 0.0001 m² )

A = 0.000314 m²

H = kAΔT / L

H = [ ( 16.5 J / s*m*K )( 0.000314 m² )( 850 K – 300 K ) ] / ( 1.5 m )

H = [ ( 16.5 J / s*m*K )( 0.000314 m² )( 550 K ) ] / ( 1.5 m )

H = 1.9 W