Rate of Temperature Increase in Liquid Heated by Radioactive Decay

To find the rate of increase in the temperature of the liquid, we first determine the number of radioactive nuclei $N(t)$ of species $X$ present at any time $t$.

Step 1: Formulate the Differential Equation for $N(t)$

The nuclide $X$ is produced at a constant rate $\alpha$ and decays at a rate proportional to the number of nuclei present, $\lambda N$. Therefore, the rate of change of $N(t)$ is given by: $\frac{dN}{dt} = \alpha - \lambda N$

Step 2: Solve for $N(t)$

Rearranging the differential equation to integrate: $\frac{dN}{\alpha - \lambda N} = dt$

Integrating both sides with the initial condition $N(0) = 0$: $\int_0^{N} \frac{dN}{\alpha - \lambda N} = \int_0^t dt$ $-\frac{1}{\lambda} \ln\left(\frac{\alpha - \lambda N}{\alpha}\right) = t$ $1 - \frac{\lambda N}{\alpha} = e^{-\lambda t}$ $N(t) = \frac{\alpha}{\lambda}\left(1 - e^{-\lambda t}\right)$

Step 3: Calculate the Decay Rate (Activity)

The activity or the number of decays occurring per unit time at time $t$ is: $\text{Activity } A(t) = \lambda N(t) = \alpha \left(1 - e^{-\lambda t}\right)$

Step 4: Calculate the Rate of Heat Supplied to the Liquid

Since each decay releases energy $E_0$, the rate of energy absorbed by the liquid (rate of heat supply $\frac{dQ}{dt}$) is: $\frac{dQ}{dt} = A(t) \cdot E_0 = \alpha E_0 \left(1 - e^{-\lambda t}\right)$

Step 5: Determine the Rate of Increase in Temperature

The heat supplied is used entirely to heat the liquid of mass $m$ and specific heat $s$: $dQ = m s \, dT \implies \frac{dQ}{dt} = m s \frac{dT}{dt}$

Equating the rate of heat supply to the thermal absorption rate: $m s \frac{dT}{dt} = \alpha E_0 \left(1 - e^{-\lambda t}\right)$

$\frac{dT}{dt} = \frac{\alpha E_0}{m s} \left(1 - e^{-\lambda t}\right)$

Thus, the correct option is A.