A village consumes 5.0×10⁴ kWh of electrical energy per day. A nuclear power plant supplying the village has an overall electrical efficiency of 30%. Assume complete fission of U-235 and take the nuclear energy released by 1.00 g of fully fissioned U-235 as 8.2×10¹⁰ J. Approximately how much U-235 must actually undergo fission each day?
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Question

A village consumes 5.0×10⁴ kWh of electrical energy per day. A nuclear power plant supplying the village has an overall electrical efficiency of 30%. Assume complete fission of U-235 and take the nuclear energy released by 1.00 g of fully fissioned U-235 as 8.2×10¹⁰ J. Approximately how much U-235 must actually undergo fission each day?

Options

G1 Physics · Accepted Answer

First convert the village's daily electrical energy to joules. 1 kWh = 3.6×10⁶ J. Therefore, E(electrical) = 5.0×10⁴×3.6×10⁶ = 1.8×10¹¹ J. The plant efficiency is 30%, so η = E(electrical) ÷ E(nuclear). Hence, E(nuclear) = 1.8×10¹¹ ÷ 0.30 = 6.0×10¹¹ J. Each gram of fully fissioned U-235 supplies 8.2×10¹⁰ J. Required mass: m = 6.0×10¹¹ ÷ 8.2×10¹⁰ ≈ 7.32 g. Therefore approximately 7.3 g of U-235 must actually fission each day. Correct option: C.