Calculating Energy Needs: Converting KWh to g/year in a Typical Home

[PhysicFailure]

A typical home uses approximately 1600 kWh of energy per month.If the energy came from a nuclear reaction, what mass would have to be converted to energy per year to meet the energy needs of the home? (in g/year)

I don't know how to do this problem :S

 

[jk4]

the things you need to know to solve the problem are:

a watt has units: J/s

E(in joules) $$= mc^{2}$$ <---this is the equation you should use to calculate how much mass is required to be converted.

$$c = 2.9979 x 10^{8}$$m/s <---speed of light.

 

[Moetasim]

I understand your confusion with this problem. Let me break it down for you.

First, let's convert the monthly energy usage of 1600 kWh to an annual usage. There are 12 months in a year, so we can simply multiply 1600 kWh by 12 to get 19,200 kWh per year.

Next, we need to convert this energy usage into a mass equivalent. This can be done using the famous equation E=mc², where E represents energy, m represents mass, and c represents the speed of light. Rearranging this equation, we get m=E/c².

To find the mass in grams, we need to convert the energy in kWh to joules (J), as the speed of light is in meters per second (m/s) and mass is typically measured in grams (g). We can do this by multiplying the kWh by 3,600,000 (1 kWh = 3,600,000 J).

So, our equation now becomes m=(19,200 kWh * 3,600,000 J/kWh)/c².

Using the value for the speed of light, c=2.998 x 10^8 m/s, we can plug in all the values and solve for mass:

m=(19,200 kWh * 3,600,000 J/kWh)/(2.998 x 10^8 m/s)^2

m=6.74 x 10^-7 g/year

Therefore, to meet the energy needs of a typical home using nuclear energy, a mass of 6.74 x 10^-7 grams would need to be converted to energy every year. This is a very small amount of mass, highlighting the immense energy potential of nuclear reactions. I hope this explanation helps you understand the problem better.