An electric heater consists of a single resistor connected across a 110 V line.

Question

An electric heater consists of a single resistor connected across a 110 V line. It is used to heat 200.0 g of water in a coffee cup from 20 C to 90 C in 2.70 min. Assuming that 90% of the energy drawn from the power source goes into heating the water, what is the resistance of the heater? Assuming no other heat losses, how much longer will it take to heat your water if you had to power the water heater with your 12.0 V car battery? An electric heater consists of a single resistor connected across a power source (v = 110 V). The heater is used to heat m = 0.2000 kg of water (c = 4186 ) from 20 degrees Celsius to 90 degrees Celsius ( T = 70 K) in t= 162 s. We will assume that only 90% of the heat dissipated by the resistor goes into heating the water. The heat required to raise the temperature of an object is equal to Q = mc T. Setting this value equal to 90% of the energy dissipated by the resistor in 162 s will allow us to calculate the resistance of the heater. We can rearrange our algebraic expression to solve for the time required to heat the water using a 12.0-V car battery, assuming everything else about the problem remains the same.

Explanation / Answer

a) amount oh heat energy required to heat the coffee cup = mCdT

= 200*4.179*(90-20) = 58506 J

power required = 58506 / (2.7*60)= 361.15 W

power from the electric heater = V^2/R

only 90% of it is used for heating

so, power utilised = 0.9*110^2/R = 361.15

R = 0.9*110^2/361.15 = 30.15 ohm

b) time required to heat if the voltage of power source is 12 V

V^2/R = 58506/t

12^2/30.15 = 58506 /t

t = 12251.25 s = 3.4 hr

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