Answer: The campus would save
$218.04.
Explanation: Total number of flourescent tubes in classroom is 12 and office is 6. The Electric Power consume by a single tube is 110W.
But,
Energy = Power * Time
For classroom energy consumed in a day that is 24hours would be
Energy = 12*110*24 = 31.7kwh
For office,
Energy = 6*110*24= 15.8kwh
Total Energy consumed at the campus for 24hrs{a full day} will be
Total Energy = 31.7kwh + 15.8kwh
= 47.5kwh.
In a year of activities at the campus given as 240days,
Energy consumed= 47.5kwh * 240
= 11400kwh.
The cost of 1kwh from the question is $0.115.
11400kwh will cost; 11400*$0.115
=$1,311
If the florescent tube are switched off for the 4hrs in a day the campus is not on session. Then we will have just 20hrs of usage in a day. So let's calculate;
Classroom,
Energy = 12*110*20= 26.4kwh
Faculty office,
Energy consumed = 6*110*20
= 13.2kwh
Total energy consumed in a day {20hrs}
= 26.4kwh + 13.2kwh= 39.6kwh.
In a year of activities at the campus given as 240days
Total Energy consumption would be
{39.6*240}kwh= 9504kwh.
The cost of 1kwh of electricity is $0.115.
Therefore, 9504kwh will cost;
$0.115 * 9504 = $1092.96.
This is more cheaper compared the the first one we calculated for 24hrs in a day.
So the campus will save;
$1,311 - $1092.96 = $218.04.
Which is the cost of energy of putting ON the flourescent tubes for 24hrs in a day minus the cost of putting them ON for just 20hrs in a day.