Volume of water:
V=50L=50×10-3m3,
Mass of water:
m=ρV=1× 103kg/ m3×50×10-3m3=50kg;
Heat absorbed by water:
Absorption Q=cm△t
=4.2 ×103J/(kg ? ℃)×50kg×(55℃-25℃)
=6.3×106J.
Answer: The water absorbed 6.3 × 106 J of heat.
Calculation of solar hot water collection area in Beijing
(1) The first question should be the heat absorbed by the water heater water during this period, and not the calorific value
< p>Absorption Q=cm (t-t0)=4.2×10?J/(kg·℃)×50kg×(60℃-20℃)=8400000J(2) The water heater radiates solar energy The efficiency of energy conversion into internal water energy is the heat absorbed by the water divided by the solar radiation energy ( the time must be the same)
The total energy of solar radiation in 4 hours is E=7000000J/h×4h=28000000J< /p>
η=Q absorption/E=8400000J/ 28000000J=30%
(3) Suppose that all the heat released by the combustion of coalwe are absorbed by water (the effectiveness is not given in the question, otherwise there is no way to do it)
Q put=Q suck
∵Q put=mq
∴m=Q put/q=8400000J/30000000J/kg=0.28kg
How many kilowatt hours of electricity are needed per ton to heat space water from 10 degrees to 55 degrees with a solar water heater? How to calculate it? Find the formula. If there is electric supplementary heating, how many kilowatt hours of electricity are required?
Define the proportion of annual investment and fuel cost savings over a certain period of time. Solar energy is the electromagnetic radiation emitted by the sun, which is captured and converted into useful energy. The calculation of hot water collection area is the proportion of annual investment and fuel cost savings over a defined period of time. The surface area of the heat absorber refers to the surface area of the inner tube of the tubeempty. The heat absorber area of each group of heat collectors is the number of vacuum tubes multiplied by the heat absorber area of each vacuum tube. .
The specific heat capacity of water is 4.2 KJ (Joules/kg degrees Celsius)
The specific heat capacity of a ton of water is 4. 2 KJ×1,000 kg=4,200
From 10 degrees Heating to 55 degrees requires thermal energy of 4,200 × (55-10) = 189,000 KJ
One kilowatt hour is equivalent at 3,600 joules
So depending on your needs, you should need 189,000 ÷ 3,600 = 52.5 kilowatt hours of electricity
Full calculation formula (the amount of electricity required to heat one ton of water from cold water temperature X to hot water temperature Y)
< p>4.2×1000×(Y-X)÷3600