A 400Wimmersion heater is placed in a pot containing 2.00 Lof water at$\mathbf{20}\mathbf{°}\mathit{C}$. (a) How long will the water take to rise to the boiling temperature, assuming that80% of the available energy is absorbed by the water? (b) How much longer is required to evaporate half of the water?

1. The power of heater is P = 400W
2. The volume of water in pot is$V=2.00Lor2\times {10}^{-3}{m}^3$

c. The temperature of water in pot is$T=20°C$

We can find heat energy absorbed by water to boil using a formula for it. Then, using the relation between power and heat energy, we can find the time taken by the water to rise to the boiling temperature assuming that the available energy is absorbed by the water. Then, using the formula for heat energy required evaporating the water and the relation between heat energy and power we can find the time required to evaporate half of the water.

Formulae:

The energy transferred by the body, $Q=mc∆T$ (i)

The mass of a body in terms of density, m = Vp (ii)

The rate of energy transfer or power generated, $P=\frac{Q}{t}$ (iii)

The heat absorbed by the body due to latent heat, $Q=m{L}_v$ (iv)

The energy absorbed by water to boil the water is given by substituting the data and equation (ii) in equation (i) as follows:

$$\begin{gathered} Q\text{'}=V\rho c∆T \\ =2\times {10}^{-3}\left(1000\right)\left(4187\right)\left(100°C-20°C\right) \\ =6.70\times {10}^{5}J \end{gathered}$$

80% of available energy is absorbed by water hence, the new power is found to be

$\begin{array}{l}P\text{'}=0.80P\\ =0.80\times 400\\ =320W\end{array}$

Thus, the time taken by water to boil is given using the above values in equation (iii) as follows:

$\begin{array}{l}t=\frac{6.70\times {10}^5\text{'}}{320}\\ =2093s\\ =34.88\mathrm{m}\\ ~35\mathrm{mm}\end{array}$

Therefore, the time taken by the water to rise to the boiling temperature assuming that 80% of the available energy is absorbed by the water is 35min.

The energyrequiredto evaporate the water is calculated using the mass value of equation (ii) in equation (iv) as follows:

$$\begin{gathered} Q=\rho V{L}_v \\ =\left(2\times {10}^{-3}\right)\left(1000\right)\left(2.256\times {10}^6\right) \\ =4.512\times {10}^{6}J \end{gathered}$$

The time required to evaporate half of the water is now given using the above power and heat of vaporization values in equation (iii) as follows:

$$\begin{gathered} t=\frac{{\displaystyle \frac{Q}{2}}}{P\text{'}} \\ =\frac{{\displaystyle \frac{4.512\times {10}^6}{2}}}{320} \\ =7.05\times {10}^{3}s \\ =117.5min \\ ~118min \end{gathered}$$

Therefore, the time required to evaporate half of the water is 118min.