Question

A surface of area $1{m^2}$ kept perpendicular to the sun rays, absorbs $1.4kJ$ of solar energy in every second. What is the amount of solar energy absorbed per second by a solar heater when it’s surface area $10{m^2}$ is exposed perpendicularly to the sun’s rays? If the efficiency of the above solar heater is 50%, then what is the amount of time taken to heat 10kg of water from $20^\circ C$ to $50^\circ C$ ?(Take $4 \cdot 2kJk{g^{ - 1}}{C^{ - 1}}$)
A) 14kJ per second, 3 minutes.
B) 4kJ per second, 3 minutes.
C) 14kJ per second, 6 minutes.
D) 4kJ per second, 6 minutes.

Answer 1

The solar energy is very useful and preferred nowadays. The more the surface area of the solar panel the more solar energy can be gathered and also the time taken for heating of the water will be less.

Complete step by step answer:
It is given that the problem has a surface area of $1{m^2}$ kept perpendicular to the sun rays, absorbs $1.4kJ$ of solar energy in every second. Here we need to find the energy absorbed per second by the solar panel if the surface area is $10{m^2}$.
As the energy absorbed by the solar panel for a surface of area $1{m^2}$ kept perpendicular to the sun rays is $1 \cdot 4kJ$ per second so for a surface area of $10{m^2}$ the energy absorbed is given by,
$\Rightarrow E = A \times 1 \cdot 4\dfrac{{kJ}}{{\sec .}}$
$\Rightarrow E = 10 \times 1 \cdot 4\dfrac{{kJ}}{{\sec .}}$
$\Rightarrow E = 14\dfrac{{kJ}}{{\sec .}}$.
So the energy absorbed for surface area $10{m^2}$ is equal to $E = 14\dfrac{{kJ}}{{\sec .}}$.
In the second part, the problem asks the time taken for 10kg water to heat from $20^\circ C$ to $50^\circ C$ if the specific heat is $4 \cdot 2kJk{g^{ - 1}}{C^{ - 1}}$ and the efficiency of the solar heater is 50%. The heat generated to heat the water from $20^\circ C$ to $50^\circ C$ is given by,
$\Rightarrow {{\eta }} = \dfrac{{{\text{output energy}}}}{{{\text{input energy}}}}$
Output energy will be the energy required for heating the water. The heating of the water is equal to,
$Q = mc\Delta T$
Where m is the mass c is the specific heat and $\Delta T$ is the change in temperature.
$\Rightarrow Q = mc\Delta T$
On substituting the corresponding values,
$\Rightarrow Q = 10 \cdot \left( {4 \cdot 2} \right)\left( {50^\circ - 20^\circ } \right)$
On simplification,
$\Rightarrow Q = 10 \cdot \left( {4 \cdot 2} \right) \cdot 30$
$\Rightarrow Q = 1260kJ$
As the input energy is the energy from the sun which is let us say for t seconds and output energy is the energy supplied to water.
$\Rightarrow {{\eta }} = \dfrac{{{\text{output energy}}}}{{{\text{input energy}}}}$
Replace the output energy as $Q = 1260kJ$ and input energy as $E = 14 \times t$.
$\Rightarrow {{\eta }} = \dfrac{{1260}}{{14 \times t}}$
On simplification,
$\Rightarrow {{\eta }} = \dfrac{{90}}{t}$
The efficiency is given as $\eta = 50\%$.
$\Rightarrow \dfrac{{50}}{{100}} = \dfrac{{90}}{t}$
$\Rightarrow 0 \cdot 50 = \dfrac{{90}}{t}$
$\Rightarrow t = \dfrac{{90}}{{0 \cdot 50}}$
$\Rightarrow t = 180s$
Converting the time in minutes.
$\Rightarrow t = \dfrac{{180}}{{60}}$
$\Rightarrow t = 3\min .$

The energy absorbed is equal to 14kJ and the time is taken to heat the water from $50^\circ$ to $20^\circ$ is equal to 3 minutes.

Note:
The efficiency of the solar panel is very important as if the efficiency of a solar panel is more then the conversion of the solar energy into electricity is more and if the efficiency of the solar panels is less then the time required to gain a certain level of voltage will take more time.