Question

The Sun delivers roughly 1 kW m−2 of power to the Earth's surface. A parabolic mirror with a 1 m radius is used to focus this energy onto a beaker containing 1 L of water. Estimate the time taken by the mirror to raise the temperature of the water by 50∘C, assuming the specific heat capacity of water to be 4200 J K−1 kg−1 and ignoring energy losses to the surroundings

Answer 1

Amount of heat required by water to increase its temperature by 50⁰C,

                        q = m c dT       -- equation 1   

Where, q= heat gained                                                                   , m= mass in kg,

c= specific heat (in terms of kJ/ kg⁰C)                       dT = temperature = increase in temperature

putting the values in equation 1

            q =1 kg x (4.2 kJ/ kg⁰C) x 50⁰C = 210 kJ

Now,

Surface area of the mirror, A = (pi) r²       , where, r = radius of the mirror = 1m

                                Or, A = 3.14 m²

1 Kw m^-2 means = 1 kJ per second per square meter surface area             ; [1 W = 1 Js^-1 ]

Amount of energy focused by mirror = incidence rate of solar energy x surface area of mirror

                                                = 1 kJ s^-1 m^-2 x 3.14 m² = 3.14 kJ s^-1

Therefore, the mirror focuses 3.14 kJ energy per second

Time required by the mirror to focus 210 kJ energy = 210 kJ / 3.14 kJ s^-1

                                                            = 66.88 seconds = 1.11 minutes