The electricity meter of a classmate's home shows 052134 at the beginning of the month and 053032 at the end of the month, How many kilowatt hours of electricity does the student's family use in a month? How many joules is it equal to? If the price of electricity per kilowatt hour is 0.50 yuan, how much electricity will his family pay this month?

The electricity meter of a classmate's home shows 052134 at the beginning of the month and 053032 at the end of the month, How many kilowatt hours of electricity does the student's family use in a month? How many joules is it equal to? If the price of electricity per kilowatt hour is 0.50 yuan, how much electricity will his family pay this month?


Electricity consumption of the month = end of the month - beginning of the month = 53032-53134 = 898 kW · H
1 kW · H = 3.6x10 ^ 6 Joule, so 898 kW · H = 898x3.6x10 ^ 6 = 3.23x10 ^ 9 Joule
Electricity charge of this month: 898 (kW · h) x0.5 (yuan / kW · h) = 449 yuan



Please design and use the electric energy meter to measure the electric power of a household appliance. Please write down the experimental instrument and measurement steps


Laboratory equipment, energy meter, clock or alarm clock, mobile phone
Step 1: turn off all the household appliances first, and record the electric energy indication and clock indication at this time
2, and then let the tested household appliances work alone for an hour (or 2 hours) to record the electric energy expression at this time
(since the unit of energy meter is kilowatt hour, the time is measured in hours.)
First, calculate the energy difference on the watt hour meter, W, then t
Formula P = w / T



The dial of an electric energy meter is marked with 1200R / (kW · h). A certain electric appliance is connected to the electric energy meter separately. When the electric appliance works for a period of time, the turntable of the electric energy meter turns 60R. If the rated voltage of the electric appliance is 220V, the passing current is 2A


After 60 R, the electric energy consumed by the appliance is 60 △ 1200 = 0.05 kW · H
The rated voltage of the electric appliance is 220 V, the working current is 2 A, so the power is 220 × 2 = 440 W
Therefore, the working time of electrical appliances is 0.05kw · h △ 440W = 0.1136h = 6.82min



The electric energy meter has been marked with "220V 10A" and "3000revs / kW · H"
The electric energy meter has been marked with the words of "220V 10A" and "3000revs / kW · H". On the circuit where it is located, how many watts should the total power of the connected electrical appliances not exceed?


10X220v=2200W