The heating element of a coffeemaker operates at 120 V and carries a current of 4.50 A. Assuming the water absorbs all of the energy converted by the resistor, calculate how long it takes to heat 0.525 kg of water from room temperature (23.0°C) to the boiling point.

It will take 313.376 sec to raise temperature to boiling point

Explanation:

We have given that potential difference V = 120 Volt

Current i = 4.50 A

So resistance

Heat flow in resistor will be equal to

It is given that this heat is used for boiling the water

Mass of the water = 0.525 kg = 525 gram

Specific heat of water 4.186 J/gram/°C

Initial temperature is given as 23°C

Boiling temperature of water = 100°C

So change in temperature = 100-23 = 77°C

Heat required to raise the temperature of water

So

t = 313.376 sec

So it will take 313.376 sec to raise temperature to boiling point

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i have found an answer to your question but use it as information.

information: because weak acids/bases have low concentration of hydrogen ions, the addition of water has a large impact on the ph. 2. a difference in 1 unit is a 10x difference in concentration for example; a liquid with ph of 3 is 10x more acidic than a liquid with a ph of 4. therefore, a liquid with a ph of 3 is stronger.

best of luck!

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