When boling water, a hot plate takes an average of 8 minutes and 41 seconds to b

Question

When boling water, a hot plate takes an average of 8 minutes and 41 seconds to boil 170 milliters of water. Assume the temperature in the lab is 79 degrees Fahrenheit. The hot plate is rated to provide 464 watts. How efficient is the hot plate? Click the icon to view conversion formulas for temperature scales. Click the icon to view the factors for conversions involving temperature differences. Click the icon to view the conversion table. Click the icon to view the density of water Click the icon to view the table of specific heats Cick the icon to view the table of specific heats. The efficiency is%. (Round your answer to three significant digits.)

Explanation / Answer

1.Answer:

"boil" has too meanings.
Normal (scientific) is to provide energy until all the water is all boiled away.
But in common use, it means to raise the temperature of the water until it just reaches the boiling point.

step 1.

Note: 1 ml of water is equivalent to 1 gram of water AND you probably remember the formula you need:

Heat Energy = Calories = Mass of water (g) X Temperature change (°C) X Calories / g C °

= 170 X 79 X 8.41/ 170 X 79 = 8.40

Step 2. NEXT calculate the amount of electrical energy delivered to the hot plate: ELECTRICAL ENERGY is measured by how much you use (WATTS) and how long you use it (TIME). SO…

______464____WATTS X __8_____MINUTES = ______3712_____ watt-minutes

NEXT convert the WATT-MINUTES into WATT-HOURS

________3712__ watt-minutes X 60 watt - minutes 1 watt - hour = ___222720______ watt-hours of electrical energy

Step 3. We now have to convert the watt-hours of electrical energy used by the hot plate in heating the water to calories of heat energy so that we can compare the energy in the water to the energy used by the hot plate.

Doing this involves a simple factor label problem when we realize that : 1 watt-hour = 860.4 calories SO….. get started –

The amount of heat energy produced by the hot plate is _______464_____

NOW WE CAN (FINALLY) CALCULATE THE EFFICIENCY OF THE HOT PLATE:

Energy supplied (the calories of heat energy produced by the hot plate) /Energy used (the calories of heat energy absorbed by the water) X 100%

= 860.4/464*100% = 1.85

2.Answer:

Energy absorved by the water = 418 X 67 X 4.18/418*67 = 41449

Energy released by the hot plate (J) = total hot plate wattage (watts) X time (s) X (1J/s / watt)

= 1550 X 20 X ( 1X418/20 / 1550 ) = 418

Efficiency = (Energy absorved by the water / energy relased by the hotplate ) 100%

= 41449 / 418 * 100% = 99%

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