THERMAL METHODS 981 



the surface according to 



x/F77^ T 2 2 



(21) 



and reaches a point 5 within the surface at a later time {h^ti). This is known as the 

 lag of the temperature wave, and Equation 22 gives the amount of time later at which 

 the minimum occurs at the depth z. 



tx — h = t = 



V^-^ = -7-^=V— (22) 



> 2 w ZyTklccTi IT 



\Jk/ca y 2 W ZyTk/ca^ IT 



A like reasoning holds for the maximum or any other phase. 

 From Equation 22, the lag of the temperature wave can be determined using the 

 above data, as follows : 



,n X 30 J 86,400 



5 = 30 cm.: t= ^/ . 





2 (.0049) 

 = 34,860 seconds 

 = 9.6 hours 



That is, the maximum or minimum (or any other phase) of temperature would 

 occur at 30 cm. below the surface, approximately 9.6 hours after its occurrence at the 

 surface. 



Equation 22 likewise gives the time it takes for the wave to travel from the surface 

 to the depth s. The apparent velocity of such a wave, which is merely the rate of travel 

 of a given maximum or minimum (or any particular phase) is then given by 



F = -^=2v/ife77^V-B- (23) 



This velocity does not express the actual speed of transmission of heat energy, which 

 is high, since that is a complex function of many variables. 



To illustrate the actual transmission of heat energy : for a good conductor, like 

 copper, at 13° C. one calorie is transmitted through one square cm. cross-section area 

 in one second, if the faces of the sample are at a temperature difference of 1° C. and 

 the width between faces is one cm. This, of course, is the definition of thermal 

 conductivity. 



An expression for the wave length may be deduced from Equation 23. 



\=:VP = 2Vk/cay/VP (24) 



Using the same data given above, the apparent velocity and wave length may be 

 readily obtained from Equations 23 and 24 respectively : 



s = 30cm.: V =^^ = 2 (.07)J-^— 

 34,860 ^ ^^86,400 



= 0.00086 cm. /sec. 



X= (0.00086) (86,400) = 2(.07) V(t) (86,400) 

 = 74.3 cm. 



