I´m preparing a camping and I´m trying to figure out how the heat a cup of tea.

Question

I´m preparing a camping and I´m trying to figure out how the heat a cup of tea. I have a rechargeable 1.2 V battery that says 2500 mAh and a tea cup that takes 300 mL. According to this the emf of the battery is 1.2 V and to find its inner resistance I measure when 2.0 A goes through the battery, then its polar Voltage (Vab) is 0.9 A, the result of that is the resistance r=0.15 ohm.

a) How much energy does the battery store according to what says on the battery ?.
b) Now I put R=0.9 ohm (heater) in to water in the tea cup and connect the battery so that current flows through the resistance. Find the Power of the heater.
c) Find the efficiency e (that is the power that is used to heat the tea water according to the total power that participate in the circuit)
d) Will the battery suffice to heat the water from 5 degree celsius to 80 degree celsius ?

Help please.

Explanation / Answer

a)

charge stored, Q = 2500 mAh = 2500*10^-3*3600 = 9000 C

So, energy stored = VQ = 1.2*9000 = 10800 J

b)

Power of the heater:

P = (1.2/(0.15+0.9))^2*0.9 = 1.18 W

c)

efficiency, e = 1.18/(1.2*(1.2/(0.15+0.9)))

= 0.861 = 86.1 percent

d)

Heat required, H = m*C*dT

C = 4186 J/kg.C

dT = 80-5 = 75 deg C

So, H = 0.3*4186*75 = 94185 J

As total energy stored is 10800 J, so it wont suffice

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