Question

g how many grams of 335u are consumed in one year, in order to supply the energy needs of a household that uses 27.0 kwh of energy per day, on the average?

Answer 1

Approximately 0.4988 grams of Uranium-335 are required to meet the annual energy needs of a household with an average usage of 27.0 kWh per day, based on the conversion of energy units and the energy density of Uranium-335.

Step-by-step explanation:

To calculate the amount of Uranium-335 consumed to supply the energy needs of a household using 27.0 kWh of energy per day, we can follow several steps:

First, we determine the total yearly energy consumption in kWh by multiplying the daily use by the number of days in a year (27.0 kWh/day × 365 days = 9855 kWh/year).

Next, we convert this energy into Joules, knowing that 1 kWh = 3.6× 106 J (9855 kWh × 3.6× 106 J/kWh

= 3.548× 1010 J).

Using the energy density of Uranium-335, which is 17 × 106 kcal/g, we convert the energy needed to kcal (3.548× 1010 J / 4184 J/kcal = 8.48× 106 kcal).

To get the mass of Uranium-335 needed, we divide the total energy needed in kcal by the energy density in kcal/g (8.48× 106 kcal / 17 × 106 kcal/g = 0.4988 g).

Thus, to meet the energy needs of a household with an average usage of 27.0 kWh per day, approximately 0.4988 grams of Uranium-335 would be consumed annually.