Chemistry, 29.10.2019 04:31 tjmoney7709

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.

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Answer from: beverlyamya

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.

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