How much electricity does it take to heat 30 liters of water to 90 degrees Celsius?

v water 30 liters 30 liters 3 * 10 (-2) cubic meters, t at the beginning of 0 degrees Celsius, at the end of t 90 degrees Celsius, water 1 * 10 3 kg cubic meters.

cWater joules (kg*C).

If the heat loss is not taken into account, the entire electrical energy is converted into heat energy and absorbed by the water, then.

e electric q water suction.

That is, the required electrical energy is eelectricity, cwater*m, water*(t, t, t), c, water*v, water*(t, t, t).

Get e joules joules.

Because 1 kWh of electricity is 1 kWh and 1000 watts * 3600 seconds joules.

Therefore, the electrical energy sought is e electricity (kWh of electricity.

w = cmδt = joules.

w (degrees.) So you need to be electric.

The amount of heat that water needs to absorb: Q=mc t=30kg

The required electrical energy is: 11340kj 3600kj kWh=

30 liters = 30l =

m=pv=1*10^3*

q=cmδt =

Suppose there is no energy loss w=q=degrees.

Without considering energy loss, about kilowatt-hour of electricity is requiredOne kilowatt-hour of electricity is 3,600 seconds * 1,000 watts = 3,600,000 joules, and the heat capacity (specific heat) of water is joules * degrees.

Temperature is a physical quantity that indicates the degree of heat and cold of an object, and microscopically speaking, it is the intensity of the thermal motion of the molecules of an object. Temperature can only be measured indirectly by certain characteristics of an object as a function of temperature, and the scale used to measure the temperature value of an object is called a temperature scale.

It specifies the starting point (zero point) of the temperature reading and the basic unit for measuring the temperature. The SI unit is the thermodynamic temperature scale (k). Other temperature scales that are more commonly used internationally are Fahrenheit (°F), Celsius (°C) and the International Practical Temperature Scale.

The specific heat capacity of water is 4200 joules (kilogram * degree), and the energy required for 1 cubic meter of water to rise from 20 degrees to 60 degrees is 4200 * 1000 * (60 20) coke coke, 1 joule 1 watt second, so the energy required is (kilowatt hours, because it is half an hour, the power is kilowatts.

Because Q=cmδT, the specific heat capacity of the hall water.

Yes, so 1 cubic meter of Sun Nao's water pants is 10 kilograms, and the heat required is 4200 1000 40 = 1680000000 joules, which is 168000000 3600000 = kWh.

How many kilowatt-hours of electricity does it take for 100 liters of water to burn from 10 degrees to 40 degrees, and if you heat it to 40 degrees, you need about 1 2 degrees of electricity, because our normal water temperature is about 20 degrees, and if you add it to 40 degrees, the electricity used at this time is relatively less, because you are 100 liters of water, and 100 liters of water is equivalent to 200 catties in 100 kilograms. Therefore, the electricity used for heating is a little more masking point, about 1 2 kWh.

If the tap water temperature is 10 degrees, the water temperature is set at 60 degrees, and the heating power of the water heater is 2000W, the time required to heat the water to the set temperature is about 2 hours, taking into account the scaling state and efficiency of the water heater itself, the time required is about 2 hours; In summer, when the tap water temperature is 20 degrees Celsius, the time fibrillation nucleus is heated to the set temperature for about an hour

The power consumption of a 60-liter water heater is 3 degrees a day, and the electricity consumption in a month is 3*30=90 degrees, if the electricity consumption of 60 liters of 10-degree cold water is heated to 60 degrees Celsius: 60*degrees, calculated according to 30 days in a month, an electric water heater.

Electricity consumption in one month: kWh. If the electricity is in yuan, the monthly electricity bill for a 60-liter water heater is 105*

Degree. According to the information of the Thermal Power Bureau, 100 liters of water need to be heated to 10 degrees, so 300 liters of water to heat 10 degrees is to use 100 liters of heat energy multiplied by 3, so destroying 300 liters of water to heat 10 degrees requires electricity.

m = 20 kg, the specific heat of ice c ice = 2100 joules kg Celsius, t1 = -10 degrees Celsius, t2 = 0 degrees Celsius, the heat of melting of ice is ice = joules kg, t3 = 30 degrees Celsius, the specific heat of water c water = 4200 joules kg Celsius.

Water at minus 10 degrees Celsius should be called ice, the heat absorbed by ice from -10 degrees Celsius to 0 degrees Celsius is Q1 = m*c ice*(t2-t1), 0 degrees Celsius ice becomes 0 degrees Celsius water and the heat absorbed is.

Q2 = m, the heat absorbed by water heating up to 30 degrees Celsius at 0 degrees Celsius is Q3=m*c water*(T3-T2).

q1=20*2100*10=joules).

q2 = 20 * joules).

q3 = 20 * 4200 * 30 = joules).

The total amount of heat absorbed is q=q1+q2+q3=joules.

If it is heated with electricity, and the heat loss is not counted, the electrical energy required is joules, i.e. kilowatt-hours.

1 kWh. Each heating rod is 1000 watts, and it can reach 50 degrees in half an hour, which is based on the wattage of the heating rod, the heat generated per minute in 300 liters of water, and the time after reaching 50 degrees is multiplied by the wattage of the heating rod, and the electricity consumption is 1 kilowatt-hour. Heating power required (kw) Weight of water (kg) Temperature difference Specific heat capacity of water {kj (kg· ) Time (seconds).