1) How much will it cost to run a 1500-watt swimming pool heater continuously for 5 hours if the utility company is charging 10 cent per kWh?
2) Determine the resistance at 20 0c, of 25m of solid aluminum round wire having a radius of 2mm. (Recall the relationship R=pl/A. The resistivity for Aluminum at 20 0c = 2.8 x 10-8 Ωm).
cost=rate*power*time= .10 (dollars/(kw hr)*1.50kw*5hr
1) To calculate the cost of running a 1500-watt swimming pool heater continuously for 5 hours, we need to follow these steps:
Step 1: Convert the power from watts to kilowatts: Since 1 kilowatt (kW) is equal to 1000 watts, we divide 1500 watts by 1000 to get 1.5 kilowatts (kW).
Step 2: Calculate the energy consumed in kilowatt-hours (kWh): Energy (in kWh) = Power (in kW) × Time (in hours). In this case, 1.5 kW × 5 hours = 7.5 kWh.
Step 3: Multiply the energy consumed by the cost per kilowatt-hour: To find the total cost, multiply the energy consumed (7.5 kWh) by the cost per kilowatt-hour (10 cents). Note that 10 cents is equivalent to $0.10. So, 7.5 kWh × $0.10 = $0.75.
Therefore, it will cost $0.75 to run a 1500-watt swimming pool heater continuously for 5 hours if the utility company is charging 10 cents per kWh.
2) To determine the resistance of 25m of solid aluminum round wire with a radius of 2mm at 20°C, we can follow these steps:
Step 1: Calculate the cross-sectional area of the wire: The formula for the area of a circle is A = πr², where "π" is pi (approximately 3.14159) and "r" is the radius. In this case, the radius is given as 2mm, which is equal to 0.002m. So, the area is A = 3.14159 × (0.002m)² = 0.000012566 m².
Step 2: Use the formula R = ρL/A to calculate the resistance, where "R" is the resistance, "ρ" is the resistivity, "L" is the length of the wire, and "A" is the cross-sectional area. The resistivity for aluminum at 20°C is given as 2.8 x 10^-8 Ωm. The length of the wire is stated as 25m.
Plugging in the values into the formula: R = (2.8 x 10^-8 Ωm) × (25m) / 0.000012566 m².
Simplifying the equation: R = (2.8 x 10^-8 Ωm) × (25m) / 0.000012566 m².
Result: The resistance of the aluminum wire at 20°C, with a radius of 2mm and a length of 25m, is calculated to be approximately 0.014 ohms.