Electrochemical Cell | 75W Bulb | 50min | 12V Battery

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SUMMARY

The discussion focuses on calculating the energy required to power a 75.0 W light bulb for 50 minutes using a 12.0 V battery. The key equation utilized is the relationship between power and energy, where 1 watt equals 1 joule per second. The solution involves converting time into seconds and applying the formula to isolate energy in joules. The final calculation confirms that 225,000 joules are needed to keep the bulb burning for the specified duration.

PREREQUISITES
  • Understanding of electrical power (Watts)
  • Basic knowledge of energy calculations (Joules)
  • Familiarity with unit conversions (minutes to seconds)
  • Concept of efficiency in electrical systems
NEXT STEPS
  • Study the relationship between power, voltage, and current in electrical circuits
  • Learn about energy consumption calculations for different electrical devices
  • Explore the concept of efficiency in electrochemical cells and batteries
  • Investigate the impact of voltage on power consumption in light bulbs
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Students studying physics, electrical engineers, and anyone interested in understanding energy consumption in electrical devices.

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Homework Statement



A 75.0-W light bulb is powered by connecting it to a standard 12.0 V auto lead-storage battery. How much energy (J) is required to keep the bulb burning for 50.0 continuous minutes ?
Assume 100 percent efficency.

Homework Equations



1 watt = 1 J/s
1 J = 1 C x 1 V

The Attempt at a Solution



(50min/12v)(60second/min)(75J/second)

I am confused about how to get Joules by itself
 
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Use the fact that 75.0 W is 75.0 J/s. If you can do something to cancel the s in 75.0 J/s, you will have the energy in Joules.
 

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