# 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...

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?

answer in Seconds

*print*Print*list*Cite

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*