Question

wind energy is becoming a more common form of energy that can be used to run household appliances. (a) if a windmill is connected to a hot water tank to heat 60 l (60 kg) of water from 15 °c to 70 °c, how much energy is required? (3 marks) (b) if energy from the utility company costs 0.35¢ per megajoule, how much money is saved in a year if the water in the tank is heated 1400 times? (3 marks) (show all your work directly below, type it in using this mathtype application.)

Explanation:

Step1: Calculate energy required per - heating

The specific heat capacity of water $c = 4.186\times10^{3}\text{ J/(kg}\cdot^{\circ}\text{C})$. The mass of water $m = 60\text{ kg}$, and the change in temperature $\Delta T=T_2 - T_1=70 - 15=55^{\circ}\text{C}$.
Using the formula $Q = mc\Delta T$, we have $Q=(60\text{ kg})\times(4.186\times10^{3}\text{ J/(kg}\cdot^{\circ}\text{C}))\times55^{\circ}\text{C}$.
$Q = 60\times4.186\times10^{3}\times55\text{ J}=60\times4.186\times55\times10^{3}\text{ J}=1.38138\times10^{7}\text{ J}$.
Since $1\text{ MJ}=10^{6}\text{ J}$, then $Q=\frac{1.38138\times10^{7}\text{ J}}{10^{6}\text{ J/MJ}} = 13.8138\text{ MJ}$.

Step2: Calculate total energy required in a year

The water is heated $n = 1400$ times a year. So the total energy $Q_{total}=nQ$.
$Q_{total}=1400\times13.8138\text{ MJ}=19339.32\text{ MJ}$.

Step3: Calculate money saved

The cost of energy from the utility company is $0.35$ cents per megajoule.
The money saved $C = 0.35\times Q_{total}$ cents.
$C=0.35\times19339.32\text{ cents}=6768.762\text{ cents}$.
Since $1$ dollar $ = 100$ cents, $C=\frac{6768.762}{100}=\$67.69$ (rounded to two decimal - places).

Answer:

(a) The energy required to heat the water once is $13.8138\text{ MJ}$.
(b) The money saved in a year is $\$67.69$.