Emporia smart plug: [ Ссылка ]
Kettle: [ Ссылка ]
Hi everyone! How’s it going? My name is Eli and in today’s video, we are going to be answering three very important questions that I know you are dying to know the answers to.
First – How much energy does it take to boil half a liter of water?
Second – How efficient is this kettle? Converting the electrical energy coming in to heat in order to boil the water.
Third – How much does it cost in electricity to boil the water?
For this experiment, we are going to add a meter that will tell us exactly how much energy we are going to be using. The meter I am using is the Emporia Smart Plug ( [ Ссылка ] ).
The first law of thermodynamics states that energy is always conserved. It can NOT be created or destroyed – only transferred. Any electrical energy will travel down the electrical cord and be transferred to heat energy and which will heat the water.
Based on the info from the Emporia app, it takes this kettle 0.0704 kWh to boil this water.
We have our first answer, so now we can move on to the next question. How efficient is this kettle? To figure this out we have to know the theoretical amount of energy required. Thankfully we have this equation that explains the heat transfer.
The change in energy is equal to the mass times the heating capacity times the change in temperature. The mass of half a liter of water is 0.5kg. The heating capacity of water is 4187 J/kg*C.([ Ссылка ]) Our change in temperature is 100 degrees Celcius (the boiling point for water) minus 19.4 C.
We can cross out our units and we are left with 168.87 KJ. We can convert KJ to kWh and get a theoretical amount of 0.047 kWh. We can compare that with our actual and get an efficiency of only 66%!!
Now the last question... How much does it cost in electricity? The rate here is $0.11/kWh so multiply that by 0.0704kWh and you get $0.0077, so less than a penny!
