# A water heater is operated by solar power. If solar collector has an area of 6.00m^2, and the...

## Question:

A water heater is operated by solar power. If solar collector has an area of 6.00m^2, and the intensity delivered by sunlight is 550W/m^2, how long does it take to increase the temperature of 1m^3 of water from 20 C to 60 C?

## Solar Water Heater:

Let the collector of solar heater have an area {eq}A {/eq}. Also let the daily intensity delivered by sunlight be {eq}I {/eq}. In addition, the heater takes time {eq}t {/eq} to increase the temperature of water of mass {eq}m {/eq} from {eq}T_1 {/eq} to {eq}T_2 {/eq}.

Therefore, the total heat energy delivered by the incident sunlight will be the heat energy absorbed by the water. Thus, we write:

$$\begin{align} IAt &= mC_w(T_2 - T_1)\\[0.3cm] t &= \frac{mC_w(T_2 - T_1)}{IA} \end{align} $$

where {eq}C_w {/eq} is specific heat of water.

## Answer and Explanation:

Given:

- Area of the solar collector: {eq}(A)= 6.00\, \mathrm{m^{2}} {/eq}

- Intensity delivered by sunlight: {eq}(I) = 550\, \mathrm{W/m^{2}} {/eq}

- Volume of water taken: {eq}(V)= 1\, \mathrm{m^{3}} {/eq}

- Density of water: {eq}(\rho) = 1000 \, \mathrm{kg/m^{3}} {/eq}

- Initial temperature of the water: {eq}(T_1) = 20\mathrm{^\circ C} {/eq}

- Final temperature of the water: {eq}(T_2) = 60\mathrm{^\circ C} {/eq}

- Specific heat of water: {eq}(C)= 4186\, \mathrm{J/kg\, ^\circ C} {/eq}

First, the mass of water is solved as follows:

$$\displaystyle{\begin{align*} m &= V\rho\\[0.3cm] &= \left(1\ \rm m^3 \right) \left(1000\ \frac{\rm kg}{\rm m^3} \right) \\[0.3cm] &= 1000\ \rm kg\\ \end{align*}} $$

Therefore, time taken by the heater in this case is given by:

$$\displaystyle{\begin{align*} t &= \frac{mC(T_2 - T_1)}{IA}\\[0.3cm] &= \frac{(1000 \ \rm kg) (4186\, \mathrm{J/kg\, ^\circ C})(60 - 20)^\circ C}{550 \ \rm J/s \cdot m^2 (6.00\ \rm m^2)} \\[0.3cm] &\approx 50,739.39\, \mathrm{s}\\[0.3cm] &\approx 14.09\, \mathrm{h} \end{align*}} \\ $$

Therefore, it takes about {eq}14.09 \ \rm hours {/eq} to increase the temperature of the water.