An aluminum rod 0.500 m in length and with a cross sectional area of 2.50 cm2 is inserted into a thermally insulated vessel containing liquid helium at 4.20 k. the rod is initially at 300 k. (a) if one half of the rod is inserted into the helium, how many liters of helium boil off by the time the inserted half cools to 4.20 k? (assume the upper half does not yet cool.) (b) if the upper end of the rod is ma

The correct choice is c

Answerthe energy required is 1.60 × 10⁻¹⁸ j, to three significant figures.explanationwe define electric potential as the ratio electric potential energy to the charge,    then the following gives us the potential difference,      the formula  may be on your formula sheet.the change in electric potential energy would be the energy that would needed to move it from point a to point b. solving for that, we get    using q = 8.00  × 10⁻¹⁹ c and  δv = 2.00 v, we get    the energy required is 1.60 × 10⁻¹⁸ j, to three significant figures.