A nuclear reactor starts producing a radioactive nuclide X from t =...

Let $N$ be the number of active nuclei at time $t$. The rate of change of the number of nuclei is given by

$\dfrac{dN}{dt} = \alpha - \lambda N$

Integrating with the initial condition $N = 0$ at $t = 0$:

$\int_{0}^{N} \dfrac{dN}{\alpha - \lambda N} = \int_{0}^{t} dt$

$-\dfrac{1}{\lambda} \ln\left(\dfrac{\alpha - \lambda N}{\alpha}\right) = t$

$N = \dfrac{\alpha}{\lambda} (1 - e^{-\lambda t})$

The rate of decay of the nuclide at time $t$ is

$A = \lambda N = \alpha (1 - e^{-\lambda t})$

Since each decay produces energy $E_0$, the rate of energy production is

$\dfrac{dE}{dt} = A E_0 = \alpha E_0 (1 - e^{-\lambda t})$

This energy is used to heat the liquid. The rate of heat absorption is

$\dfrac{dQ}{dt} = m s \dfrac{dT}{dt}$

Equating the rate of energy production to the rate of heat absorption:

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

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

Answer: $\dfrac{\alpha E_0}{m s}(1 - e^{-\lambda t})$