Question

A surface of area $1m^2$ kept perpendicular to the sun rays, absorbs $1.4kJ$ of solar energy in every second.
(a) What is the amount of solar energy absorbed per second by a solar heater when its surface of area $10m^2$ is exposed perpendicularly to the sun's rays?
(b) If the efficiency of the above solar heater is 50%, then what is the amount of time taken to heat $10kg$ of water from 20 C to 50 C? (Take $4.2kJ{kg}^{-1}{C}^{-1}$)

• A. 14 kJ per second, 3 minutes
• B. 4 kJ per second, 3 minutes
• C. 14 kJ per second, 6 minutes
• D. 4 kJ per second, 6 minutes

Answer 1

The correct option is A 14 kJ per second, 3 minutes

(a) The amount of solar energy absorbed by a surface kept perpendicular to sun's rays in one second is $=1.4kJ{m}^{-2}$.
For 10 metre square area, solar energy absorbed $=1.4\times 10=14kJ$ per second.
(b) The efficiency of the solar heater is $=50$%
The amount of thermal energy required for $10kg$ of water to raise its temperature from 20 C to 50 C is $=Q=ms\Delta t$
$\Rightarrow Q=10kg\times 4.2kJ{kg}^{-1}{C}^{-1}\times 30C=1260kJ$
Energy received by solar heater in time ${}^{\prime }{t}^{\prime }$ is $=1.4\times {10}^3\times 10\times t$
But 50% of $1.4\times {10}^3\times 10\times t=1260\times {10}^3$
$\therefore t=180s=3$ minutes