How long will it take to increase the temperature of 1.0 m3 of water from 29°C to 60°C?
A hot-water heater is operated by solar power. If the solar collector has an area of 6.0 m2 and the power delivered by sunlight is 550 W/m2, how long will it take to increase the temperature of 1.0 m3 of water from 29°C to 60°C?
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The hot-water heater is operated by solar power. The solar collector has an area of 6.0 m^2 and the power delivered by sunlight is 550 W/m^2. The time it takes to increase 1 m^3 of water from 29°C to 60°C has to be determined.
The total heat collected is the sunlight falling on an area of 6.0 m^2. This is equal to 550*6 = 3300 W or 3300 J/s.
It takes 1 liter of water (assuming a density of 1 kg/l) 4181 J of heat to raise by 1 degree C. 1 m^3 of water is equal to 1000 l . The rise in temperature is 60 - 29 = 31 C. The heat required for this is 4181*31*1000 = 129611 kJ. This amount of heat is collected in 129611*10^3/3300 s = 39.27*10^3 s
It takes 39.27* 10^3 s to collect the heat required to raise 1 m^3 of water from 29 C to 60 C.
Area of heating surface = 6.0 m2
Power absorbed from sun light = 550 W/m2 = 550 J/s/m2
Rate of power absorption of the solar water heater = 6.0*550 J/s
= 3300 J/s
Quantity of Water = 1.0 m3
Mass of 1.0 M2 of water = 1000 Kg
Initial temperature = 29 degree C
Final Temperature = 60 degree C
Specific heat of water = 4186 J/Kg/degree C
Heat required to raise the temperature of 1000 Kg of water from 29 degree C to 60 degree C = 4168*(60-29)*1000 J = 1.292*10^8 J
Time required = 1.292*10^8 / 3300 / 3600 Hours = 10.9 Hours
Time required to raise the temperature of 1.0 M3 of water from 29 degree C to 60 degree C = 10.9 Hours
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