At 7.65 cents per kilowatt-hour, what does it cost to operate a 10.0 hp motor for 6.00 hr?
At 7.65 cents per kilowatt-hour, what does it cost to leave a 75.0 W light burning 24.0 hours a day?
cost=rate*time*power
in the first, cnvert hp to kwatts
in the second, convert W to kw.
To calculate the cost of operating a motor or a light, we need to convert the power input into kilowatt-hours (kWh) and then multiply it by the cost per kilowatt-hour.
1) Cost to operate a 10.0 hp motor for 6.00 hours:
Step 1: Convert horsepower (hp) to kilowatts (kW).
1 hp = 0.746 kW (approximately)
So, 10.0 hp = 10.0 * 0.746 kW = 7.46 kW
Step 2: Calculate the energy consumed in kilowatt-hours (kWh).
Energy (kWh) = Power (kW) * Time (hours)
Energy (kWh) = 7.46 kW * 6.00 hours = 44.76 kWh
Step 3: Calculate the cost.
Cost = Energy (kWh) * Cost per kWh
Cost = 44.76 kWh * $0.0765/kWh = $3.43 (rounded to two decimal places)
Therefore, the cost to operate a 10.0 hp motor for 6.00 hours at a rate of 7.65 cents per kilowatt-hour is approximately $3.43.
2) Cost to leave a 75.0 W light burning for 24.0 hours a day:
Step 1: Convert watts (W) to kilowatts (kW).
1 kW = 1000 W
So, 75.0 W = 75.0 / 1000 kW = 0.075 kW
Step 2: Calculate the energy consumed in kilowatt-hours (kWh).
Energy (kWh) = Power (kW) * Time (hours)
Energy (kWh) = 0.075 kW * 24.0 hours = 1.80 kWh
Step 3: Calculate the cost.
Cost = Energy (kWh) * Cost per kWh
Cost = 1.80 kWh * $0.0765/kWh = $0.1377 (rounded to four decimal places)
Therefore, the cost to leave a 75.0 W light burning for 24.0 hours a day at a rate of 7.65 cents per kilowatt-hour is approximately $0.1377.
To calculate the cost of operating a motor or a light, we need to follow these steps:
1. Convert the given power to kilowatts (kW). One horsepower (hp) is equivalent to 0.7457 kilowatts (kW). So, for the first question, we convert 10.0 hp to kW: 10.0 hp * 0.7457 kW/hp = 7.457 kW. For the second question, the power is already given in watts (W), so we don't need to convert.
2. Multiply the power (in kW) by the time (in hours) to get the energy consumed (in kilowatt-hours, kWh). For the first question, the motor operates for 6.00 hours, so the energy consumed is 7.457 kW * 6.00 hr = 44.742 kWh. For the second question, the light burns for 24.0 hours a day, so the energy consumed is 75.0 W * 24.0 hr = 1800 Wh, which we also need to convert to kilowatt-hours by dividing by 1000: 1800 Wh / 1000 = 1.8 kWh.
3. Finally, multiply the energy consumed (in kWh) by the cost per kilowatt-hour to get the total cost. For both questions, the cost per kilowatt-hour is given as 7.65 cents. To convert cents to dollars, we divide by 100, so the cost per kilowatt-hour is 7.65 cents / 100 = $0.0765. Now, we can calculate the cost for each question:
- For the first question: Total cost = 44.742 kWh * $0.0765/kWh = $3.426
- For the second question: Total cost = 1.8 kWh * $0.0765/kWh = $0.1377
So, it would cost $3.426 to operate a 10.0 hp motor for 6.00 hours, and it would cost $0.1377 to leave a 75.0 W light burning 24.0 hours a day.