Chapter 18, Problem 62GP

(a)

To determine

The monthly bill.

Answer to Problem 62GP

  $\$44.10permonth$

Explanation of Solution

Given:

Power of heater, $1.8kW,3.0h/day$

Power of lightbulb, $100W,6.0h/day$ and number of bulbs: $4$

Power of electric stove: $2.0kWh/day$

Formula used:

To calculate the cost of the energy, multiply the kilowatts of power consumed by the number of hours in operation times the cost per $kWh$ .

Calculation:

According to the question, first calculate the total power:

  $\left(1.8kW\right)\left(3.0h/d\right)+4\left(0.1kW\right)\left(6.0h/d\right)+\left(3.0kW\right)\left(1.4h/d\right)+\left(2.0kWh/s\right)=14.0kWh/d$

Calculate the cost of the energy, $$

  $Cost=\left(14.0kWh/d\right)\left(30d\right)\left(\frac{\$0.105}{kWh}\right)=\$44.10permonth$

Conclusion:

The monthly bill: $\$44.10permonth$

(b)

To determine

Mass of coal that must be burned by $35\%$ efficient power plant to provide the yearly needs of this house hold.

Answer to Problem 62GP

Approximately $1800kgofcoal$ .

Explanation of Solution

Given:

Efficiency is $35\%$ .

Energy of coal is $7000kcal/kg$

Formula used:

Formula used to calculate coal mass:

  $Coalmass=\frac{Household}{\left(Eff.\right)\left(Coalenergypermass\right)}$

Calculation:

calculate coal mass by using the formula:

  $Coalmass=\frac{Household}{\left(Eff.\right)\left(Coalenergypermass\right)}$

  $Coalmass=\frac{\left(14.0kWh/d\right)\left(365d\right)\left(\frac{1000W}{kW}\right)\left(\frac{3600s}{1h}\right)}{\left(0.35\right)\left(7000\frac{kcal}{kg}\right)\left(\frac{4186J}{1kcal}\right)}=1794kg\approx 1800kgofcoal$

Conclusion:

The coal mass is approx. $1800kgofcoal$ .