Physicist Making Tea

$P=\frac{E}{t}=\frac{C\times100\times250}{95}=VI$ Substituting in $C=4.2,V=230$ , we can readily determine $I$ to be 4.805 Amperes.

We know that heat gainer is given by $\text{IVt}$ and $\text{mc}{\triangle}_{\text{t}}$ .
Equating both the equations, we get:-
$\begin{aligned} \text{mc}{\triangle}_{\text{t}} & = \text{IVt}\\ 250 × 4.2 × (100-0) & = \text{I} × 230 × 95\\ \text{I} & = \dfrac{250×4.2×100}{230×95}\\ & = \color{#3D99F6}{\boxed{4.805 \text{A}}} \end{aligned}$

$H$ $=$ $m$ $c$ $\Delta$ $\theta$ where $\Delta$ $\theta$ is the change in the temperature; $m$ is the mass of water (in this case); $H$ is the amount of heat (basically energy) and $c$ is the specific heat capacity

Also, $H$ $=$ $V$ $i$ $t$ where $V$ is the voltage supplied; $i$ is the current passing through coil and $t$ is the time taken

Equating these two expressions and solving for $i$ we get $i$ $\approx$ $4.805$

$Q=Pt\\ mc\Delta\theta = IVt\\ 250(4.2)(100-0)=I(230)(1 \times 60 + 35)\\ I=\dfrac{250(4.2)(100)}{230(95)}\approx \boxed{4.805\text{A}}$