.During a lightning flash, there exists a potential differenceof V cloud - V gro

Question

.During a lightning flash, there exists a potential differenceof Vcloud - Vground = 1.2× 109 V between a cloud and the ground. As aresult, a charge of -25 C is transferred from the ground to thecloud. A) how much work Wground-cloud is done on the charghe by theelectric force? B) If the work done by the electric force were used toaccelerate a 1100kg automobile from rest, what would be its finalspeed? C) If the work done by the electric force were converted intoheat, how many kilograms of water at 0 degree C could be heated to100 degree C? .During a lightning flash, there exists a potential differenceof Vcloud - Vground = 1.2× 109 V between a cloud and the ground. As aresult, a charge of -25 C is transferred from the ground to thecloud. A) how much work Wground-cloud is done on the charghe by theelectric force? B) If the work done by the electric force were used toaccelerate a 1100kg automobile from rest, what would be its finalspeed? C) If the work done by the electric force were converted intoheat, how many kilograms of water at 0 degree C could be heated to100 degree C?

Explanation / Answer

(a) The amount of work Wground-cloud is done on thecharghe by the electric force is                   Wground-cloud = Vq                                         = (1.2*109V)(25C)                                         = 30.0*109J (b) Initial speed (u) of the automobile = 0 Let v be the final speed of the automobile. Mass (m) of the automobile = 1100 kg Then according to the work-energy theorem, the workdone by theautomobile is equal to the change in its kinetic energy. so W = KEfinal - KEinitial           =(1/2)mv2 - (1/2)mu2          =(1/2)mv2 - 0          =(1/2)mv2       v = (2W / m) Here W = 30.0*109J          m = 1100kg (c) Specific heat of water is c = 4186 J/kg.Co Temperature difference is T = 100oC We know that Q = mcT                       m = Q / (cT) Here Q = 30.0*109J                   =(1/2)mv2 - 0          =(1/2)mv2       v = (2W / m) Here W = 30.0*109J          m = 1100kg (c) Specific heat of water is c = 4186 J/kg.Co Temperature difference is T = 100oC We know that Q = mcT                       m = Q / (cT) Here Q = 30.0*109J         

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