91Ó°ÊÓ

1. The efficiency of the overall system is $\mathrm{Î\mu }=20\%$
2. The mass of the water is$m=200\text{Lor}200×{10}^3\text{g}$
3. The initial temperature of the water is$T_i=20\text{°°ä}$
4. The final temperature of the substance is$T_f=40.0\text{°°ä}$
5. The time for raising the temperature is$t=1.0\text{hor}3600\text{s}$
6. The intensity of the incident sunlight is $I=700{\text{W/m}}^{\text{2}}$

A substance's specific heat capacity is the amount of energy required to raise its temperature by one degree Celsius. We can use the concept of specific heat of the water and find the energy collected by the solar heater. We can use the expression of the power consumed by the system and then use the expression of the intensity of the incident sunlight and find the area of the collector of the heater.

Formulae:

The heat energy required by a body, $Q=mc\mathrm{Δ}T$ …(¾±)

Where,m = mass

c = specific heat capacity

$\mathrm{Δ}T$= change in temperature

Q= required heat energy

The power generated by a body,$P=\frac{Q}{\mathrm{Î\mu }t}$ …(¾±¾±)

The intensity of the heat generated,$I=\frac{P}{A}$ …(¾±¾±¾±)

The specific heat of the water is$c=4.18J/g^{0}C$

The expression of the power consumed by the system is given by substituting the value of equation (i) in equation (ii) as follows:

$\begin{array}{c}P=\frac{cm(T_f−T_i)}{\mathrm{Î\mu }t}\\ =\frac{4.18{\text{J/g}}^{\text{0}}\text{C}×200×{10}^3\text{g}(40.0\text{°°ä}−20\text{°°ä})}{20\%×3600\text{s}}\\ =2.3×{10}^4\text{W}\end{array}$

The collected area of the intensity of the incident sunlight using equation (iii) and above values is given as:

$\begin{array}{c}A=\frac{2.3×{10}^4\text{W}}{700{\text{W/m}}^{\text{2}}}\\ =33{\text{m}}^{\text{2}}\end{array}$

Hence, the value of the collected area is $33{\text{m}}^{\text{2}}$