Select the correct answer. if two half-lives have passed since a scientist collected a 1.00-gram sample of u-235, how much u-235
1 answer:
You might be interested in
Answer:
Q(total) = 283Kj
Explanation:
5 Heat Transitions …
Specific Heats => c(s) = 0.50cal/g∙⁰C, c(l) = 1.0 cal/g∙⁰C, c(g) = 0.48 cal/g∙⁰C
Phase Transition Constants => ΔHᵪ = Heat of Fusion = 80 cal/g; ΔHᵥ = Heat of Vaporization = 540cal/g
Note => Phase change regions => no temp. change occurs when 2 phases are in contact (melting and evaporation). Only when single phase substance exists (s, l or g) does temperature change occur. See heating curve for water diagram. The increasing slopes are temperature change regions and heat flow is given by Q =mcΔT. The horizontal slopes are phase changes ( melting & evaporation) and heat flow for each of those regions is given by Q = m·ΔH. Each transition energy is calculated individually (see below) and added to obtain the total heat flow needed.
Q = mcΔT for temperature change regions of the heating curve (single phase only)
Q = m∙ΔH for phase transition regions of the heating curve (2 phases in contact)
Solid (ice) => Melting Pt => Q(s) = mcΔT = (90.5g)(0.50cal/g∙⁰C)(11⁰C) = 478 cal
Melting (s/l) => Liquid (water) => Q(s/l) = m∙ΔHᵪ = (90.5g)(80cal/g) = 7240 cal
Liquid (water) => Boiling Pt => Q(l) = mcΔT = (90.5g)(1.0cal/g∙⁰C)(100⁰C) = 9050 cal
Boiling (l/g) => Gas (steam) => Q(l/g) = m∙ΔHᵥ = (90.5g)(540cal/g) = 48,870 cal
Gas (steam) => Steam @ 145⁰C => Q(g = mcΔT = (90.5g)(0.48cal/g∙⁰C)(45⁰C) = 2036 cal
Total Heat Transfer (Qᵤ) = Q(s) + Q(s/l) + Q(l) + Q(l/g) + Q(g)
= 478cal +7240cal + 9050 cal + 48,870cal + 2036cal
= 67,674 cal x 4.184 j/cal = 283,148 joules = 283 Kj
