While completing your homework, imagine the study space you are using employs a space heater with a rating of 30 kW, a refrigerator (for snacks) that is rated for 5 kW, and a light fixture with two 100 W bulbs, all operating at a constant rate for the 1.5 h it takes you to finish.
A. What is the total power consumed at any given moment?
B. How much energy do you consume to complete the task?
C. Assume the utility company charges 6.5 cents per unit of energy in part b, how much did it cost you to complete the assignment?
D. If you were willing to pay the utility company $5, what would be the power rating of an additional appliance you could add to your study space?

a) Let suppose that all appliance work simultaneously. The total power consumed ([tex]\dot W[/tex]), measured in kilowatts, is the sum of the power of the space heater, the refrigerator and the two light bulbs:

[tex]\dot W = 30\,kW+5\,kW+0.2\,kW[/tex]

[tex]\dot W = 35.2\,kW[/tex]

The total power consumed at any given moment is 35.2 kilowatts.

b) A common unit to quantify the energy consumption ([tex]E[/tex]) is the kilowatt-hour, which is obtained by multiplying the total power consumed, measured in kilowatts, by the operation time ([tex]\Delta t[/tex]), measured in hours. If we know that [tex]\dot W = 35.2\,kW[/tex] and [tex]\Delta t = 1.5\,h[/tex], then the energy consume to complete the task is:

[tex]E= \dot W \cdot \Delta t[/tex] (1)

[tex]E = (35.2\,kW)\cdot (1.5\,h)[/tex]

[tex]E = 48.75\,kWh[/tex]

The energy consumed to complete the task is 48.75 kilowatt-hours.

c) The total cost ([tex]C[/tex]), measured in US dollars, is determined by multiplying the energy consumed by the unit cost ([tex]c[/tex]), measured in US dollars per kilowatt-hour, That is: ([tex]c = 0.065\,\frac{USD}{kWh}[/tex], [tex]E = 48.75\,kWh[/tex])

[tex]C = c\cdot E[/tex] (2)

[tex]C = \left(0.065\,\frac{USD}{kWh} \right)\cdot (48.75\,kWh)[/tex]

[tex]C = 3.168\,USD[/tex]

The cost for completing the assignment is 3.168 US dollars.

d) If we know that [tex]C = 5\,USD[/tex], [tex]c = 0.065\,\frac{USD}{kWh}[/tex] and [tex]\Delta t = 1.5\,h[/tex], then the energy and power consumed are, respectively:

[tex]E = \frac{C}{c}[/tex]

[tex]E = \frac{5\,USD}{0.065\,\frac{USD}{kWh} }[/tex]

[tex]E = 76.923\,kWh[/tex]

[tex]\dot W = \frac{E}{\Delta t}[/tex]

[tex]\dot W = \frac{76.923\,kWh}{1.5\,h}[/tex]

[tex]\dot W = 51.282\,kW[/tex]

And the additional appliance is:

[tex]\Delta \dot W = 51.282\,kW-35.2\,kW[/tex]

[tex]\Delta \dot W = 16.082\,kW[/tex]

16.082 kilowatts of power rating could be added to the study space.