A water is operated by solar power. If the solar collector has an area of 5.25m2, and the intensity delivered by sunlight is 475W/m2, how long does it take to increase the temperature of .600m3 of water from 20.C to 85.0C?
To calculate the time it takes to increase the temperature of the water, we need to determine the amount of heat energy required and then divide it by the rate at which the solar collector can deliver that energy.
First, let's calculate the amount of heat energy required to heat the water:
Mass of water (m) = volume × density
Density of water = 1000 kg/m^3 (approx.)
Volume of water (V) = 0.600 m^3
So, the mass (m) = V × density = 0.600 m^3 × 1000 kg/m^3 = 600 kg
The specific heat capacity of water (c) is 4186 J/kg°C.
Change in temperature (ΔT) = final temperature - initial temperature
= 85.0°C - 20.0°C = 65.0°C
The amount of heat energy required (Q) can be calculated using the formula:
Q = m × c × ΔT
Now, let's substitute the values:
Q = 600 kg × 4186 J/kg°C × 65.0°C
Calculating Q, we get:
Q = 16,305,600 J
Now, let's determine the power delivered by the solar collector:
Intensity of sunlight = 475 W/m^2
The power (P) delivered on the solar collector is calculated by multiplying the intensity by the area:
P = intensity × area
P = 475 W/m^2 × 5.25 m^2
Calculating P, we get:
P = 2493.75 W
Now, to find the time (t) it takes to heat the water, we divide the energy required by the power delivered:
t = Q / P
Substituting the values:
t = 16,305,600 J / 2493.75 W
Calculating t, we get:
t = 6549.02 seconds
Therefore, it takes approximately 6549 seconds to increase the temperature of 0.600 m^3 of water from 20.0°C to 85.0°C using solar power.