Question

How much time would it take for 1 litre of water to boil, initially at $0^\circ {\rm{C}}$, if a concentrator type solar heater of area $5\;{{\rm{m}}^2}$ is used, which can reflected $90\%$ of energy incident on it?
(Take solar constant as $1.4\;{\rm{kW}}{{\rm{m}}^{ - 2}}$ ).
(a) 1.11 min
(B) 2.22 min
(C) 0.56 min
(D) 1.86 min

Answer 1

To determine the required time, you have to calculate the total amount of energy required to boil 1 liter of water and calculate the energy absorbed by the water due to the heater. Then, the division of the total energy required and the energy absorbed will give you the required time of boiling.

Complete step by step answer:
It is given that the amount of water is 1 litre and its initial temperature is $0^\circ {\rm{C}}$. The solar heater area is $5\;{{\rm{m}}^2}$ and the heater can reflect $90\%$ of energy incident on it. Take the values of the solar constant $1.4\;{\rm{kW}}{{\rm{m}}^{ - 2}}$ in the calculations.
Let the required time of boiling is $t$ , boiling temperature of the water is $100^\circ {\rm{C}}$ and specific heat of the water is $4.2 \times \;{\rm{J/g}}^\circ {\rm{C}}$
The total solar amount of energy falling on the $1\;{{\rm{m}}^2}$ area of the solar heart is,
${Q_h} = {K_s} \times 1\;{\rm{kJ}}$
Substitute the value of solar constant in the above equation.
Therefore, we get
$\begin{array}{l} {Q_h} = 1.4\; \times 1\;k{\rm{J}}\\\ {{\rm{Q}}_h} = 1400\;{\rm{J}} \end{array}$
Now for $5\;{{\rm{m}}^2}$ area of the heater, the total amount of energy falling on the solar heater is,
$\begin{array}{l} {Q_h} = 1400\;{\rm{J}} \times 5\\\ {Q_h} = 70000\;{\rm{J}} \end{array}$
The heater can reflect $90\%$ of the energy incident on it, so the amount of energy absorbed by the water per second from the heater is,
$\begin{array}{l} {Q_h} = 7000\;{\rm{J/s}} \times \dfrac{{90}}{{100}}\\\ {Q_h} = 6300\;{\rm{J/s}} \end{array}$…… (1)
By using the mass of the water, specific heat and temperature difference of the water, the calculation of the energy required to boil 1 litre of water is
$Q = mc\Delta T$…… (2)
Here, $m$ is the mass of the water, $c$ is the specific heat of the water and $\Delta T$ is the temperature difference of the water.
To calculate the temperature difference, use the initial and boiling temperature of the water. Therefore we get
$\Delta T = {T_2} - {T_1}$
Here, ${T_2}$ is the boiling temperature of the water and ${T_1}$ is the initial temperature of the water.
Substitute the values in the above equation.
Therefore, we get
$\begin{array}{l} \Delta T = 100^\circ {\rm{C}} - 0^\circ {\rm{C}}\\\ \Delta T = 100^\circ {\rm{C}} \end{array}$
Now substitute the values of the $m$, $c$ and $\Delta T$ in equation (2).
Therefore, we get
$\begin{array}{l} Q = 1\;\;{\rm{l}} \times \dfrac{{1000\;{\rm{g}}}}{{1\;{\rm{l}}}} \times {\rm{4}}{\rm{.2}}\;{\rm{J/g}}^\circ {\rm{C}} \times {\rm{100}}^\circ {\rm{C}}\\\ Q = 4.2 \times {10^5}\;{\rm{J}} \end{array}$…… (3)
From equation (1) and (3), the time required to boil the water is,
$t = \dfrac{Q}{{{Q_h}}}$
Substitute the values in the above equation.
Therefore, we get
$\begin{array}{l} t = \dfrac{{4.2 \times {{10}^5}\;{\rm{J}}}}{{6300\;{\rm{J/sec}}}}\\\ t = 66.66\;{\rm{sec}} \times \dfrac{{1\;{\rm{min}}}}{{60\;{\rm{sec}}}}\\\ t = 1.11\;{\rm{min}} \end{array}$
Hence, option (A) is correct, that is $1.11\;{\rm{min}}$.

Note: Try to remember the expression of the amount of heat required to boil the water, the value specific heat of the water and boiling temperature of the water. Also, convert the total amount of incident energy upto 90% as it is given in the question that the heater can reflect $90\%$ of energy incident.