Respuesta :
The total cost of the energy consumption for the given time period is $0.0364
The given parameters include;
- power rating of the heater, P = 4 kW = 4000 W
- time taken to completely boil the water,[tex]t = 4 \min = 4 \min \times \frac{1 \ h}{60 \ \min} = 0.067 \ h[/tex]
- mass of the water, m = 1 kg
- initial temperature of the water, t₁ = 20 ⁰C
- final temperature of the water, t₂ = 50 ⁰C
- specific heat capacity of water, c = 4200 J/kg⁰C
- cost of electrical energy = $0.12
The total energy consumed in the process of heating the water;
Total electrical Energy = Energy of the heater (kWh) + Energy of the heated water (kWh)
Energy of the heater (kWh) = power x time
= 4 kW x 0.067 h
= 0.268 kWh
Energy of the heated water (J) = mcΔt
where;
Δt is the change in temperature = t₂ - t₁
= 50 ⁰C - 20 ⁰C = 30⁰C
Energy of the heated water (J) = 1 x 4200 x 30
= 126,000 J
Electrical power of the heated water (Watts) = [tex]\frac{126,000 \ J}{4 \ \min} \times \frac{1 \min}{60 \ s} = 525 \ W[/tex]
Energy of the heated water (kWh) = [tex]\frac{525}{1000} \times 4\min\times \frac{1 \ hr}{60 \min} = 0.035 \ kWh[/tex]
Total electrical Energy = 0.268 kWh + 0.035 kWh
= 0.303 kWh
The total cost of the energy consumed is calculated as;
[tex]Total \ cost = \frac{\$ \ 0.12}{kWh} \times 0.303 \ kWh = \$ \ 0.0364[/tex]
Therefore, the total cost of the energy consumption is $0.0364
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