Question

A radionuclide with disintegration constant $\lambda$ is produced in a reactor at a constant rate $\alpha$ nuclei per sec. During each decay energy $E_0$ is released. 20% of this energy is utilised in increasing the temperature of water. Find the increase in temperature of m mass of water in time t. Specific heat of water is S. Assume that there is no loss of energy through water surface.

Answer 1

$\frac{0.2 \alpha E_0}{m S} \left( t - \frac{1}{\lambda} (1 - e^{-\lambda t}) \right)$

Answer 2

Let $N(t)$ be the number of nuclei of the radionuclide at time $t$.
The rate of production of nuclei is given as $\alpha$.
The rate of disintegration of nuclei is $\lambda N(t)$.
The rate of change of the number of nuclei is given by the differential equation:
$\frac{dN}{dt} = \alpha - \lambda N$

Assuming that at $t=0$, there are no nuclei of the radionuclide, i.e., $N(0)=0$.
We solve the differential equation:
$\frac{dN}{dt} + \lambda N = \alpha$
This is a first-order linear differential equation. The integrating factor is $e^{\int \lambda dt} = e^{\lambda t}$.
Multiply by the integrating factor:
$e^{\lambda t} \frac{dN}{dt} + \lambda N e^{\lambda t} = \alpha e^{\lambda t}$
$\frac{d}{dt}(N e^{\lambda t}) = \alpha e^{\lambda t}$
Integrate from $t'=0$ to $t'=t$:
$\int_0^t \frac{d}{dt'}(N(t') e^{\lambda t'}) dt' = \int_0^t \alpha e^{\lambda t'} dt'$
$[N(t') e^{\lambda t'}]_0^t = \alpha \left[ \frac{e^{\lambda t'}}{\lambda} \right]_0^t$
$N(t) e^{\lambda t} - N(0) e^0 = \frac{\alpha}{\lambda} (e^{\lambda t} - e^0)$
Since $N(0)=0$:
$N(t) e^{\lambda t} = \frac{\alpha}{\lambda} (e^{\lambda t} - 1)$
$N(t) = \frac{\alpha}{\lambda} (1 - e^{-\lambda t})$

The rate of decay at time $t$ is $R(t) = \lambda N(t)$.
$R(t) = \lambda \frac{\alpha}{\lambda} (1 - e^{-\lambda t}) = \alpha (1 - e^{-\lambda t})$ nuclei per second.

Each decay releases energy $E_0$. The rate of energy release at time $t$ is $P(t) = R(t) E_0$.
$P(t) = \alpha E_0 (1 - e^{-\lambda t})$ joules per second.

The total energy released from $t=0$ to time $t$ is the integral of the power over this time interval:
$E_{released} = \int_0^t P(t') dt' = \int_0^t \alpha E_0 (1 - e^{-\lambda t'}) dt'$
$E_{released} = \alpha E_0 \int_0^t (1 - e^{-\lambda t'}) dt'$
$E_{released} = \alpha E_0 \left[ t' - \frac{e^{-\lambda t'}}{-\lambda} \right]_0^t$
$E_{released} = \alpha E_0 \left[ t' + \frac{1}{\lambda} e^{-\lambda t'} \right]_0^t$
$E_{released} = \alpha E_0 \left( (t + \frac{1}{\lambda} e^{-\lambda t}) - (0 + \frac{1}{\lambda} e^0) \right)$
$E_{released} = \alpha E_0 \left( t + \frac{1}{\lambda} e^{-\lambda t} - \frac{1}{\lambda} \right)$
$E_{released} = \alpha E_0 \left( t - \frac{1}{\lambda} (1 - e^{-\lambda t}) \right)$

20% of this energy is utilised in increasing the temperature of water. The energy absorbed by water is $Q$.
$Q = 0.20 \times E_{released} = 0.20 \alpha E_0 \left( t - \frac{1}{\lambda} (1 - e^{-\lambda t}) \right)$

This energy $Q$ increases the temperature of mass $m$ of water with specific heat $S$. The relationship between heat absorbed and temperature change is $Q = m S \Delta T$, where $\Delta T$ is the increase in temperature.
$m S \Delta T = 0.20 \alpha E_0 \left( t - \frac{1}{\lambda} (1 - e^{-\lambda t}) \right)$

Solving for $\Delta T$:
$\Delta T = \frac{0.20 \alpha E_0}{m S} \left( t - \frac{1}{\lambda} (1 - e^{-\lambda t}) \right)$

This can also be written as:
$\Delta T = \frac{0.2 \alpha E_0}{m S \lambda} (\lambda t - (1 - e^{-\lambda t}))$
$\Delta T = \frac{0.2 \alpha E_0}{m S \lambda} (\lambda t - 1 + e^{-\lambda t})$