A pot containing 10 kg of water at 20 o C is heateduntil all the water is at 80

Question

A pot containing 10 kg of water at 20oC is heateduntil all the water is at 80oC, by placing a resistor(having resistance R = 930 ) in the water and running acurrent through it. What current must be placed through theresistor if the 10 kg of water is to be raised from 20oCto 80oC in 5 minutes? Assume that water has aspecific heat of c = 4186 J/(kg * oC), which is theenergy needed to heat one kilogram of water by1oC. a. 2.0 A b. 3.0 A c. 4.0 A d. 5.0 A e. 6.0 A A pot containing 10 kg of water at 20oC is heateduntil all the water is at 80oC, by placing a resistor(having resistance R = 930 ) in the water and running acurrent through it. What current must be placed through theresistor if the 10 kg of water is to be raised from 20oCto 80oC in 5 minutes? Assume that water has aspecific heat of c = 4186 J/(kg * oC), which is theenergy needed to heat one kilogram of water by1oC. a. 2.0 A b. 3.0 A c. 4.0 A d. 5.0 A e. 6.0 A

Explanation / Answer

What a question. On paper it's fine but in realitynahhh! first thing we notice it that we're working in Joules andtime. Put them together and you get joules/second which is anotherway of saying power. There are a couple of different equations forpower which, if you start with P=IV, you can put togetherusing V=IR, they are P=IV , P=V2/R,P=I2R. If you wanted to raise the temperature of 10kg of water from20oC to 80oC, how much energy isrequired? (10kg)(4184J/(kg*oC)(60oC)=2510400J Ok, now we know that we're doing it over the course of 5minutes. So we've got the energy and we've got the time; let's dothe power. P=IV=I2R=(2510400J)/(300s)=8368J/s I2R=8368J/s I2(930)=8368J/s I2=8.998J/s I=3.0A Yay!!!

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