In a certain city, electricity costs $0.16 per kw⋅h. what is the annual cost for electricity to power a lamppost for 7.00 h per day with a 100. w incandescent light bulb versus an energy efficient 25 w fluorescent bulb that produces the same amount of light? assume 1 year=365 days.

So, you can save $ 40.89 - $ 10.22 = $ 30.67 per each lampost if you replace the incadescent bulb with the energy efficient fluorescent bulb.

Explanation:

kW.h means kilowatt.hour and is a unit of energy derived from the definition of power.

The definition of power is energy per time:

Thus, you get:

For the units, you get:

1. Calculate the energy used by the 100 W incandescent light bulb that is used 7.00 h per day, in a year (365 days).

2. Calculate the energy used by the 25 W flourescent bulb:

3. Now, you can calculate both costs:

    100 W bulb:

     25 W bulb:

when 25.0 ml of 0.700 mol/l naoh was mixed in a calorimeter with 25.0 ml of 0.700 mol/l hcl, both initially at 20.0 °c, the temperature increased to 22.1 °c. the heat capacity of the calorimeter is 279 j/°c.

the equation for the reaction is:

naoh + hcl → nacl + h₂o

moles of hcl = 0.0250 l hcl ×

0.700

mol hcl

1 l hcl = 0.0175 mol hcl

volume of solution = (25.0 + 25.0) ml = 50.0 ml

mass of solution = 50.0 ml soln ×

1.00

ml soln

= 50.0 g soln

δt=t2-t1

= (22.1 – 20.0) °c = 2.1 °c

the heats involved are

heat from neutralization + heat to warm solution + heat to warm calorimeter = 0

q1+q2+q3=0

nδh+mcδt+cδt=0

0.0175 mol ×

δh

+ 50.0 g × 4.184 j·g⁻¹°c⁻¹ × 2.1 °c + 279 j°c⁻¹ × 2.1 °c = 0

0.0175 mol ×

δh

+ 439.32 j + 585.9 j = 0

0.0175 mol ×

δh

= -1025.22 j

δh=1025.22j0.0175 mol= -58 600 j/mol = -58.6 kj/mol