capacitor after a time equal to 1 tau? Problems A 200-W heater is used to heat w

Question

capacitor after a time equal to 1 tau? Problems A 200-W heater is used to heat water in a cup. It consists of a single resistor R placed across 110V. Find R Assume that 90 percent of the energy goes into heating the water. How long does it take to heat 0.25 kg of water from 15 to 100degree C? How long does it take to boil this water away after it reaches 100degree C? A variable resistance R is connected across a potential difference V that remains constant. When R has the value R1, the current is 6.0 A. When R is increased by 10 Ohm from this value, the current drops to 2.0 A. Find R1 and V.

Explanation / Answer

Given Power of the heater , P = 200W potential difference , V = 110 V a) Resistance can be calculated using the relation     P = V^2 / R 200W = (110 V)^2 / R     R = 60.5 b) Mass of water , m = 0.25 kg     Change in temperature, T = (1000C - 150C) = 850C     Since E = m Cw  T    where E(90%) is the amount of heat required and    Cw is the specific heat of water, Cw = 4186 J/kg0C     And P = 0.9 E / t             t = 0.9 ( m Cw  T ) / P =0.9 [0.25kg *(4186 J/kg0C) * 850C] / 200 W             t = 400.286s   or 7.07 min c) Time taken to boil the water after it reaches 1000C is           t = 0.9 E / P = 0.9 *m Lv / P where Lv is the latent heat of vaporization,Lv = 2260 *10^3 J /kg           t = 0.9 *0.25kg * 2260 *10^3 J /kg / 200W           t = 2542.5 s or 42.37 min           t = 0.9 *0.25 *

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