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February 28, 2015

February 28, 2015

Posted by **physics.....** on Tuesday, February 19, 2008 at 2:54pm.

- physics. -
**drwls**, Tuesday, February 19, 2008 at 3:02pmThere are two ways to do this and they should agree, if all of the electrical power goes into heating the water. Let's see if they do.

Power consumed by the "thingy" is 1500 J/s * 3.0 min * 60 s/min = 270,000 J

Q = Change of water's internal energy

= m C *delta T

= 1500 g* (1.0 cal g/C)*41 C *4.18 J/cal= 257,070 J

The latter number is more reliable. The 1500 W rating of the heater may have not been quite been reached due to low voltage, or heat transferred to other regions

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