Question

The sun delivers roughly 1kWm2 of power to the Earth’s surface. A parabolic mirror with a 1.5 m radius is used to focus this energy onto a beaker containing 2 L of water. Estimate the time taken by the mirror to rise the temperature of water by 60°C, assuming Cp,m= 75291 J K mol and ignoring energy losses to the surroundings.

Answer 1

Explanation:

The amount of heat required by water to increase its temperature by $60^{o}C$,

                       q = $m \times c dT$ .............. (1)    

where,       q = heat gained  

                 m= mass in kg,

                  c = specific heat (in terms of $kJ/ kg^{o}C$)                                                          

                dT = increase in temperature

Now, putting the given values into equation ( 1) as follows.

                 q = $m \times c dT$

                     = $1 kg \times (75.291 kJ/ kg^{o}C) \times 60^{o}C$

                    = 4517.46 kJ

Now,  surface area of the mirror, A = $\pi \times r^{2}$

where,            r = radius of the mirror

                         = 1.5 m

Therefore,     area =  $\pi \times r^{2}$

                              = $3.14 \times (1.5 m)^{2}$

                              = 7.065 $m^{2}$

As, 1 $Kw m^{2}$ means = 1 kJ per second per square meter surface area.                  (as 1 W = 1 J/s)

Hence, amount of energy focused by mirror = incidence rate of solar energy x surface area of mirror

                    = $1 kJ/s m^{2} \times 3.14 m^{2}$

                   = 3.14 kJ s-1

Therefore, the mirror focuses 3.14 kJ energy per second

Time required by the mirror to focus 4517.46 kJ  energy = $\frac{4517.46 kJ }{3.14 kJ/s}$

                          = 1438.68 seconds

or,                     = $\frac{1438.68 sec}{60 sec/min}$

                         = 23.97 min

Thus, we can conclude that the time taken by the mirror to rise the temperature of water is 23.97 min.