-Math in the electric kettle

abstract

In the process of learning mathematics, I found an interesting phenomenon: in life, mathematics is everywhere. Whether at home or at school, whether it is household appliances or school desks and chairs, as long as you carefully observe and think, you will find that they all contain mathematical problems. Even if a matchbox is dropped, you can still find that it is related to the famous mathematical theorem-Pythagorean theorem, so there is another solution to Pythagorean theorem.

Electric kettle; utilize energy

In the cold winter, people always like to drink hot water, which is not only warm, but also good for their health. So, in the office, we can often see such a scene: a person comes to the office, puts a small pot of water in the hot pot, boils it, pours it into his own cup, and goes to work. Then, another man came to the office and put a small pot of water in the hot pot like the first man.

This can't help but make me think: Is it more energy-saving to boil a pot of water at one time or to boil a pot of water twice separately? So, I launched the following research:

First, I used an electric kettle marked "220 V ~1350hz" to boil different amounts of water several times (maximum capacity of electric kettle 1.8L, minimum capacity of 0.2L) and recorded the time. The results are as follows:

① ② ③ ④ ⑤

Water volume (liter)1.81.41.00.60.2

It takes 8 minutes, 30 seconds, 5 minutes, 45 seconds, 4 minutes, 56 seconds, 3 minutes, 53 seconds, 2 minutes and 55 seconds.

Solution: ∫U = 220v.

I= 13A

∴P=UI=220U× 13A=2860W

(1) t = 8 minutes 30 seconds =5 10 seconds.

w = Pt = 2860 w×5 10s = 1458600j

② T = 5 minutes and 45 seconds =345 seconds

At this rate, it will take 3 105/7 seconds to boil 1.8L of water.

w = Pt = 2860 w×3 105/7≈ 12686 14.3j

③ T = 4 minutes and 56 seconds =296 seconds

At this speed, it takes 532.8 seconds to boil 1.8L of water.

w = Pt = 2860 w×532.8 = 1523808j

④ T = 3 minutes and 53 seconds =233 seconds

At this rate, it takes 699 seconds to boil 1.8L of water.

w = Pt = 2860 w×699 = 1999 140j

⑤ t = 2 minutes and 55 seconds = 175 seconds

At this rate, if you want to boil 1.8L of water, it will take 1575 seconds.

W=Pt=2860W× 1575=4504500J

∫4504500j > 1999 140j > 1523808j > 12686 14.3j > 1458600j

Boil a pot of water at a time to save electricity.

I come to a conclusion: the more water you boil at a time, the higher the utilization rate of electric energy.

In order to save energy and reduce emissions, we can't boil water and save electricity at one time like the people in the office mentioned above. If everyone can do this, the effect of saving electricity will be amazing. Take Wenzhou alone:

There are about 7 million people in Wenzhou, and every three people can be regarded as a family. Every family has an electric kettle (maximum volume 1.8L), and needs 1 kettle of water every day, which is equivalent to 0.6L of water per person every day. If you burn 1.8L water at a time instead of 0.6L water at a time, the electricity saved can be calculated as follows:

Power consumption of three-time combustion: t=3 minutes and 53 seconds ×3=699 seconds.

w = Pt = 2860 w×699 = 1999 140j

Power consumption for once boiling: t=8 minutes and 30 seconds =5 10 second.

w = Pt = 2860 w×5 10s = 1458600j

The electricity saved by a family in one day:1999140j-1458600j = 540540j.

1* * Family: 7 million ÷3≈2333333.

All households can save electricity one day: 2333333× 540540 =1261259819820j.

There are about 7 million people in Wenzhou, and every 9 people can be regarded as an office. Each office has an electric kettle (maximum volume 1.8L), which requires 3 pots of water every day, equivalent to 0.6L of water per person per day. If you burn 65,438+0.8L water at a time instead of 0.6L water at a time, the electricity saved can be calculated as follows:

Electricity consumption for each person to boil water by himself: t=3 minutes and 53 seconds ×9=2097 seconds.

W=Pt=2860W×2097=5997420J

Power consumption for burning a three-mouth boiler: t=8 minutes and 30 seconds ×3= 1530 seconds.

w = Pt = 2860 w× 1530s = 4375800j

Electricity saved by an office in one day: 5997420J-4375800J =1621620J.

I have an office: 7 million ÷ 9 ? 777778.

Power saving for all offices in one day: 777778×1621620 =1261260360360j.

If you only count the electric kettles in the office and at home, you can save electricity in Wenzhou in one day:

1261260360360+1259819820 = 252252018018065438 ≈ 700700 kwh = 7000 kwh.

Then in one year, Wenzhou can save electricity: 700,700× 365 = 255,755,500 degrees.

But soon, I found a new problem: if you boil too much water at a time, the water will cool down after a long time. For an office without a thermos, the cooled water is equivalent to not boiling, so it is not only a waste of electricity, but also a waste of electricity. So, I began to study the problem of water cooling.

I boil 1.8L water and record the temperature every two minutes. The results are as follows:

Time (minutes) 0 2 4 6

Temperature (℃) 100 8 1 75 70

Time (minutes) 8 10 12 14

Temperature (℃) 66 62 59 57

Time (minutes) 16 18 20 22

Temperature (℃) 54 565 438+0 49 47

Time (minutes) 24 26 28 30

Temperature (℃) 45 44 42 40

After doing experiments with water at various temperatures, I found that water at 75℃ tastes very hot and boiling, so you can drink it without being too hot. It can be called hot water, while water at 54℃ tastes warm and can be called warm water. In order to meet people's needs in winter, we call water below 54℃ cold water, so 1.8L water is boiling 65438.

It is found that 1.8L water is enough for three people to drink.

It is observed that only a few people come to the office half an hour before working hours, and most people come to the office about ten minutes before working hours.

So, we can draw the conclusion that:

People who come to the office half an hour before work time (about 1~2 people) can burn enough water for their own consumption; People who come to the office ten minutes before work time (about 7-8 people) boil as much water at a time as the electric kettle does at a time. Because it is the peak time for people to come to the office, you can drink the remaining water before it cools down. In this way, the power saved can be calculated as follows:

Everyone only burns their own water consumption: t=3 minutes and 53 seconds ×9=2097 seconds.

W=Pt=2860W×2097=5997420J

Reasonable water consumption: t=3 minutes 53 seconds× 3+8 minutes 30 seconds× 2 =1719 seconds.

w = Pt = 2860 w× 17 19 = 49 16340j

You can save electricity at the end of the day: 5997420J-4916340J =1081080J.

Although compared with the original, it saves less electricity:1621620j-1081080j = 540540j.

However, there are two pieces of 0.6L water, which need not be heated again because the temperature has dropped, wasting electricity:

T=3 minutes 53 seconds ×2=466 seconds

W=Pt=2860W×466= 1332760J

So this not only saves electricity, but also saves more electricity at the end of the day: 1332760J-540540J = 79220J.

Summary: To study the problem of electric kettle, we should not only consider the power consumption of boiling water, but also consider the problem of water cooling. Of course, for practical application, it is more important to make a concrete analysis in combination with the actual situation. In order to save limited energy and make better use of energy, we should learn to use our mathematical knowledge to analyze the problems in life, so that mathematics can be integrated into our lives and make greater contributions to us.