An oven is plugged into a 120 V electrical outlet and "cooks" the food by using a 144 ohm resistor light bulb that converts electrical energy to both light and heat energy. If the light bulb converts 75% of the electrical energy to thermal energy, and the oven is used for a total of 11.6 hours during one month, to the nearest hundredth of a megajoule, how much thermal energy was produced?
To find the thermal energy produced by the light bulb, we need to calculate the electrical energy consumed by the oven and then multiply it by the efficiency factor.
To calculate the electrical energy consumed by the oven, we can use the formula:
Electrical energy (E) = Power (P) × Time (t)
First, we need to calculate the power used by the oven. We can use Ohm's law to find the power:
Power (P) = (Voltage)^2 / Resistance
Given:
Voltage (V) = 120 V
Resistance (R) = 144 ohms
Plugging in the values, we get:
Power (P) = (120 V)^2 / 144 ohms = 100 W
Next, we can calculate the electrical energy consumed by the oven over a month.
Given:
Time (t) = 11.6 hours
Converting the time to seconds:
Time (t) = 11.6 hours × 3600 seconds/hour = 41760 seconds
Now, we can calculate the electrical energy consumed:
Electrical energy (E) = Power (P) × Time (t) = 100 W × 41760 s = 4176000 J
Since the light bulb converts 75% of the electrical energy to thermal energy, we need to multiply the electrical energy by the efficiency factor:
Thermal energy produced = Electrical energy × Efficiency
Thermal energy produced = 4176000 J × 0.75 = 3132000 J
Finally, we can convert joules to megajoules by dividing the thermal energy by 1,000,000:
Thermal energy produced = 3132000 J / 1,000,000 = 3.13 MJ (to the nearest hundredth)
Therefore, the thermal energy produced by the oven during one month is approximately 3.13 megajoules.
To find the thermal energy produced, we can use the formula:
Thermal Energy = Power × Time
First, we need to calculate the power used by the light bulb. The power can be calculated using Ohm's Law:
Power (P) = (Voltage (V))^2 / Resistance (R)
Given:
Voltage (V) = 120 V
Resistance (R) = 144 ohm
Power (P) = (120 V)^2 / 144 ohm
Power (P) = 100 W
The light bulb converts 75% of electrical energy to thermal energy. Therefore, the effective power used for thermal energy conversion is:
Effective Power = 75% of Power (P)
Effective Power = 0.75 × 100 W
Effective Power = 75 W
Now we can calculate the thermal energy produced:
Thermal Energy = Effective Power × Time
Given:
Time = 11.6 hours
Thermal Energy = 75 W × 11.6 hours
To convert hours to seconds:
1 hour = 3600 seconds
Thermal Energy = 75 W × 11.6 × 3600 s
Now, we need to convert the thermal energy from Joules to Megajoules by dividing by 1,000,000:
Thermal Energy = (75 × 11.6 × 3600) / 1,000,000 Megajoules
Calculating the final result:
Thermal Energy = 299.88 / 1,000,000 Megajoules
Thermal Energy ≈ 0.00030 Megajoules (to the nearest hundredth of a Megajoule)
Therefore, approximately 0.00030 Megajoules of thermal energy were produced.