A 2.5 kg block of water at 52°C is to be heated to 100°C using a 240 V, 2.0 A immersion heater. Calculate the time taken to heat the water. Assume the specific heat capacity of water is 4.2 x 10³ Jkg⁻

The initial temperature is $52 \, ^\circ C$ and the final temperature is $100 \, ^\circ C$.

$$\Delta T = T_{final} - T_{initial} = 100 \, ^\circ C - 52 \, ^\circ C = 48 \, ^\circ C$$

Since a change in Celsius is equivalent to a change in Kelvin, $\Delta T = 48 \, K$.

The mass of the water is $m = 2.5 \, kg$, and the specific heat capacity of water is $c = 4.2 \times 10^3 \, Jkg^{-1}K^{-1}$.
The heat energy $Q$ is given by the formula $Q = mc\Delta T$.

$$Q = (2.5 \, kg) \cdot (4.2 \times 10^3 \, Jkg^{-1}K^{-1}) \cdot (48 \, K)$$ $$Q = 2.5 \times 4.2 \times 48 \times 10^3 \, J$$ $$Q = 10.5 \times 48 \times 10^3 \, J$$ $$Q = 504 \times 10^3 \, J$$ $$Q = 5.04 \times 10^5 \, J$$

The immersion heater is rated at $I = 2.0 \, A$ and $V = 240 \, V$.
The electrical power $P$ is given by the formula $P = VI$.

$$P = (240 \, V) \cdot (2.0 \, A)$$ $$P = 480 \, W$$

Assuming all the electrical energy is converted into heat energy, the heat energy supplied by the heater is $Q = P \cdot t$, where $t$ is the time taken.
Rearranging for $t$:

$$t = \frac{Q}{P}$$ $$t = \frac{5.04 \times 10^5 \, J}{480 \, W}$$ $$t = \frac{504000 \, J}{480 \, W}$$ $$t = 1050 \, s$$

$$t = 1.05 \times 10^3 \, s$$

Comparing this with the given options:

(a) $1.05 \times 10^3 \, s$

(b) $1.05 \times 10^3 \, s$

(c) $1.05 \times 10^2 \, s$

(d) $0.105 \times 10^4 \, s = 1.05 \times 10^3 \, s$

Options (a), (b), and (d) are numerically identical. We will select option (a).

The final answer is $\boxed{\text{A. } 1.05 \times 10^3 \text{s}}$.