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Rate of temperature rise of a liquid heated by a radioactive nuclide produced at a constant rate

A reactor makes nuclide X at a constant rate alpha; its decays heat a liquid. The nuclei build up to alpha/lambda, so the heating starts at zero and levels off: dT/dt = (alpha E0/ms)(1 - e^-lambda t) - option A, found by fitting the options to the thermometer's measured rate.

Sub-topics covered

• Radioactive decay law: activity = lambda N
• Production at a constant rate with decay: dN/dt = alpha - lambda N
• Saturation: N -> alpha/lambda, activity -> alpha
• Energy released per decay heating a liquid: ms dT/dt = E0 lambda N
• Decay heat after production stops

What it teaches

• dN/dt = alpha - lambda N with N(0) = 0 gives N = (alpha/lambda)(1 - e^-lambda t)
• The activity lambda N = alpha(1 - e^-lambda t) starts at zero and rises to alpha, when decay balances production
• With no heat loss ms dT/dt = E0 lambda N, so dT/dt = (alpha E0/ms)(1 - e^-lambda t) - option A
• Check every candidate at t = 0 (nothing has decayed, so no heating) and as t -> infinity (activity = alpha): only A passes both
• Traps: (B) grows without limit; (C) puts lambda where the production rate belongs; (D) heats at full power before any nucleus exists
• After the reactor stops, the nuclei present keep decaying: decay heat falls as e^-lambda t

Simulation ID

• ADV-2026-P2-PHY-Q02 · revision 1

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