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April 20, 2014

April 20, 2014

Posted by **Sasami** on Thursday, December 3, 2009 at 12:23am.

(a) If the temperature of the water is 21.6 ° C before you turn on the heater, how long does it take for the water temperature to reach 100 ° C after you turn on the heater? Provide your answers in minutes.

1 min

(b) After the water temperature reached 100 ° C, if you let the water keep boiling with the heater turn on for 20 minutes, how much water will be left in the container? Latent heat, or the heat of vaporization of water is 2260 J/g. Assume that all the vaporized water is removed from the container.

Last Question I will ask for the semester. Please help.

- AP Physics -
**drwls**, Thursday, December 3, 2009 at 5:27am(a) You want to heat the water

delta T = 100-21.6 = 78.4 degrees.

The energy needed is

Q = C M *delta T

= 4.19*280*78.4 = 91,980 joules

Power*Time = 91,980

Time (in seconds) = 91,980/293 = 314 s

Convert that to minutes.

(b) After that length of time, additional heat added is used to vaporize. In 20 minutes, the added heat is 1200 seconds * 293 J/s = 351,600 J.

Divide that by the heat of vaporization to get the number of grams of water vaporized. Subtract that from 280 g to get the number of grams left.

- AP Physics -
**Sasami**, Thursday, December 3, 2009 at 6:03amThank you Dr. DRWLS

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