You go to kitchen to boil water. You pour 280 gram of water to a container, and put the container on a 293 W electric heater. Assume that all the heat from the heater is used to heat the water. The specific heat of water is 4.19 J/g ° C. As always, be accurate to 4 significant digits.

Question

(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.

Answer 1

(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.

Answer 2

Thank you Dr. DRWLS