You find yourself the victim of a first world problem: your cell phone has died.

Question

You find yourself the victim of a first world problem: your cell phone has died. You plug It into a wall outlet to charge. The battery in your phone has a volume of 17.875 cm^3, a density of 4.9 g/cm^3, a charging efficiency of 95%, and a specific energy of 0.55 MJ/kg. Assuming the wall outlet is putting out 10W of power, how long will it take your phone to charge? 4098 s 90 min there is no sufficient information to calculate the answer 1542 s 5074 s 65 min 9875 s how much energy was dissipated as heat during the charging process? 2540 kJ 2540 J 9276 J 8976 kJ 8976 J 8763 J the answer cannot be obtained with the available information Calculate the change in temperature of the battery assuming that: the battery Is an Isolated system it is entirely made of LiCoO2 its specific heat is c_p = 125 J/mol*K 6 K 87 K 5.6 K 12K 22.7K 1.34 K 9- 98K After a long time (assuming you kept your phone isolated), you measure that the temperature of the screen of the phone has increased by 12K (i.e., Delta T = 12K). What is the value of the heat capacity (in units of J/K) of the rest of the phone (i.e., everything except the battery). 12 J/K 99.8 J/K 901 J/K 21.3 J/K 641 J/K 111.9 J/K

Explanation / Answer

1. Mass = densityxvolume=17.878x4.9gm=87.6022gm=.0876022Kg

Energy needed=.55MJ*.0876022MJ=.04818MJ

Time taken for 100% efficiency= Total energy/Power=.04818*10^6/10=4818secs

Efficiency=95%

Net time taken approx: (100/95)*4818 ~5072 secs

Ans: 5074 secs

b) Net energy needed: 100*.04818MJ/95 = .0507MJ

Energy wasted as heat: ~ .0507-.04818MJ ~ 2540J

c) Weight of battery: 87.6022gm

No of moles:n= 87.6022/97.87=.895 where 97.87gm is the molar weight

Change in temperature: 2540/(n*Cp)=22.7K

d) Considering the battery, the energy needed to increase its Temperature by 12K is

H=.895*125*12=1342.6J

For the rest of the phone, Cp=H/delT =1342/12=111.9 J/K

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