An electric iron rated at 750 W is used for 8 hours per day. How much electrical energy (in units, where 1 unit = 1 kWh) is consumed by the iron in one day?

Introduction / Context:
Electricity bills for households and industries are usually based on energy consumption measured in kilowatt hour units, often simply called units. Electrical appliances such as irons, heaters, and fans have a rated power in watts and are used for certain durations each day. This question asks you to calculate the daily energy consumption of an electric iron with a known power rating and usage time. It tests your understanding of the relationship between power, time, and electrical energy in practical units.

Given Data / Assumptions:

• Power rating of the iron P = 750 W.
• Usage time per day t = 8 hours.
• 1 kilowatt hour (kWh) is defined as the energy used by a 1 kW appliance running for 1 hour.
• 1 unit of electrical energy on the bill is equal to 1 kWh.

Concept / Approach:
Electrical energy consumed is equal to power multiplied by time. To express energy in kWh, we convert power into kilowatts and time into hours, then multiply. The formula in billing units is Energy (in kWh) = Power (in kW) * Time (in hours). Here, 750 W equals 0.75 kW, and the iron is used for 8 hours. So the daily energy consumption is 0.75 * 8 kWh. This gives a straightforward numerical answer that can be compared with the options provided.

Step-by-Step Solution:
Step 1: Convert the power from watts to kilowatts: 750 W = 0.75 kW.Step 2: Note the usage time per day: t = 8 hours.Step 3: Use the formula Energy (kWh) = Power (kW) * Time (h).Step 4: Substitute the values: Energy = 0.75 * 8 = 6 kWh.Step 5: Recognise that 1 kWh is one electrical unit, so the iron consumes 6 units of energy in one day.

Verification / Alternative check:
You can verify the calculation using joules if desired. One kilowatt hour equals 1000 W * 3600 s = 3.6 * 10^6 J. The iron uses 750 W for 8 hours, which is 8 * 3600 = 28800 seconds. The energy in joules is E = 750 * 28800 = 21.6 * 10^6 J. Dividing by 3.6 * 10^6 J per kWh gives E = 6 kWh. This matches the previous result, confirming that 6 units is correct. This cross check using fundamental units ensures that no conversion errors have occurred.

Why Other Options Are Wrong:
Option B, 600 units, would correspond to 600 kWh, which is unrealistic for a single 750 W appliance used only 8 hours per day; it would require much higher power or usage time. Option C, 0.6 units, is too small and would come from treating 750 W as 0.075 kW or using 0.8 hours instead of 8 hours. Option D, 60 units, is ten times too large and would require either 10 times the usage time or power. Only option A, 6 units, fits the correct calculation based on the given power and time.

Common Pitfalls:
Common errors include forgetting to convert watts to kilowatts, mixing up hours and minutes, or misplacing decimal points when multiplying. Some students also confuse power and energy, thinking that the wattage itself is the energy consumed. To avoid mistakes, always convert to kilowatts, use time in hours, and apply the simple formula Energy = Power * Time for kWh calculations. Writing each step clearly, as shown here, helps prevent unit and decimal errors in exam situations.

Final Answer:
6 units