An electric heater consists of a single resistor connected across 110 V. it is u

Question

An electric heater consists of a single resistor connected across 110 V. it is used to heat 230.0 g of water in a coffee cup from 20° C to 90"Cin2.70min. Star Cu 1) Assuming that 90% of the energy drawn from the power source goes into heating the water, what is the resistance of the ?Use c 4186 J/kg-K (Express your answer to two significant figures.) Pr 26 You currently have 1 submissions for this question. Only 3 submission are allowed You can make 2 more submissions for this question. Your submissions: 26 Computed value: 26Submitted: Thursday, March 22 at 1:39 PM Feedback: Right answer! St 2) 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? (Express your answer to two significant figures) 3.3 h Submit You currently have 2 submissions for this question. Only 3 submission are allowed. You can make 1 more submissions for this question. Your submissions: 3.3X Computed value: 3. 4 1uomitted: Thursday, March 22 at 2.37 PM itted: Thursday, March 22 at 2:37 PM Feedback:

Explanation / Answer

1) Total energy delivered to the water = (90-20)*0.23*4186 = 67394.6 J

This energy is delivered to water in 2.7 minutes i.e. 162 sec

So the power delivered to the water is 67394.6 / 162 = 416 W

But this is only 90% of power output of heater

So the output power of heater is 416*100 / 90 = 462.2 W

We know that power = V^2 / R

So 110^2 / R = 462.2

which gives R = 26.18ohm

2) Since we know from 1) above that the given amount of water will require 67394.6 joules.

and R was calculated to be 26.18 ohm

So the power output from 12V car battery will be V^2 / R = 5.5 W

So total time required to heat the water = 67394.6 / 5.5 = 12253 sec = 204.23 minutes = 3.4 hours

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