Question

In a solar water heater, energy from the Sun is gathered by water that circulates through tubes in a rooftop collector. The solar radiation enters the collector through a transparent cover and warms the water in the tubes; this water is pumped into a holding tank. Assume that the efficiency of the overall system is$\mathbf{20}\mathit{\%}$ (that is, $\mathbf{80}\mathit{\%}$of the incident solar energy is lost from the system).What collector area is necessary to raise the temperature of$\mathbf{200}\mathbf{}\mathbf{\text{L}}$ of water in the tank from $\mathbf{\text{20°C}}$to$\mathbf{40}\mathbf{\text{°C}}$ in$\mathbf{1}\mathbf{.0}\mathbf{}\mathbf{\text{\hspace{0.17em}h}}$ when the intensity of incident sunlight is $\mathbf{700}\mathbf{}{\mathbf{\text{W/m}}}^{\mathbf{\text{2}}}$?

Short answer

The collector area of the solar water heater is $33{\text{m}}^{\text{2}}$

Step-by-step solution

The given data

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

Understanding the concept of specific heat

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, $Q=mc\Delta T$ …(i)

Where,m = mass

c = specific heat capacity

$\Delta T$= change in temperature

Q= required heat energy

The power generated by a body,$P=\frac{Q}{\epsilon t}$ …(ii)

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

Calculation of the collected area

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:

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

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

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

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