A nuclear reactor starts producing a radioactive nuclide X from t = 0, at a constant rate of per second. Each decay of X produces energy E_0, which is utilized to heat a liquid of mass m and specific heat s. Assuming no heat loss from the liquid and taking as the decay constant of X, the rate of increase in the temperature of the liquid is:
A. E_0 m s (1 - e^ - t )
Let N be the number of active nuclei at time t. The rate of change of the number of nuclei is given by dN dt = - N Integrating with the initial condition N = 0 at t = 0: _ 0 ^ N dN - N = _ 0 ^ t dt - 1 ( - N ) = t N = (1 - e^ - t ) The rate of decay of the nuclide at time t is A = N = (1 - e^ - t ) Since each decay produces energy E_0, the rate of energy production is dE dt = A E_0 = E_0 (1 - e^ - t ) This energy is used to heat the liquid. The rate of heat absorption is dQ dt = m s dT dt Equating the rate of energy production to the rate of heat absorption: m s dT dt = E_0 (1 - e^ - t ) dT dt = E_0 m s (1 - e^ - t ) Answer: E_0 m s (1 - e^ - t )
Related: Physics — Nuclear Physics · All PYQ Banks