A 800-w iron is left on the ironing board with the base exposed to the air. About 85 percent of the heat generated by iron dissipate through its base with surface area of 0.15 ft2 . Calculate the amount of heat the iron dissipates in 2 hours, the heat flux on the surface of iron and total cost of electricity wasted in 2 hours if unit cost of electricity to be $0.11kwh.

Answer:

ΔQ = 4896 KJ

Heat Flux = 57407.55 W/m²

Cost = $ 0.176

Explanation:

The amount of heat dissipated in two hours can be given as:

[tex]Heat\ Dissipation = \Delta Q = Pt\eta[/tex]

where,

P = Power of Iron = 800 W

t = time taken = 2 hrs

η = Conversion efficiency = 85% = 0.85

[tex]\Delta Q = (800\ W)(7200\ s)(0.85)\\\\[/tex]

ΔQ = 4896 KJ

Now, for heat flux:

[tex]Heat\ Flux = \frac{P}{Surface\ Area}\\\\Heat\ Flux = \frac{800\ W}{0.15\ ft^{2}}\frac{1\ ft^{2}}{0.092903\ m^{2}}[/tex]

Heat Flux = 57407.55 W/m²

Now, for total heat energy in KWhr in two hours:

[tex]Heat\ Energy = (Power)(Time)\\\\Heat\ Energy = (0.8\ KW)(2\ hr)\\\\Heat\ Energy = 1.6\ KWh[/tex]

Now, for the cost:

[tex]Cost = (Heat\ Energy)(Unit\ Cost)\\Cost = (1.6\ KWh)(\$ 0.11\ /KWh)[/tex]

Cost = $ 0.176