As illustrated in the figure, the 2.0 resistor is immersed in a...

The equivalent resistance of the parallel combination of the $6\text{ }\Omega$ and $2\text{ }\Omega$ resistors is:

$R_p = \dfrac{6 \times 2}{6 + 2} = 1.5\text{ }\Omega$

The total resistance of the circuit is:

$R_{eq} = 1 + R_p = 1 + 1.5 = 2.5\text{ }\Omega$

The total current drawn from the battery is:

$I = \dfrac{V}{R_{eq}} = \dfrac{6}{2.5} = 2.4\text{ A}$

The voltage across the parallel combination is:

$V_p = I \times R_p = 2.4 \times 1.5 = 3.6\text{ V}$

The power dissipated by the $2\text{ }\Omega$ resistor is:

$P = \dfrac{V_p^2}{R} = \dfrac{(3.6)^2}{2} = 6.48\text{ W}$

The heat produced in the $2\text{ }\Omega$ resistor in $15$ minutes ($900$ seconds) is:

$H = P \times t = 6.48 \times 900 = 5832\text{ J}$

Using the heat capacity $C = 2000\text{ J K}^{-1}$, the rise in temperature $\Delta T$ is:

$\Delta T = \dfrac{H}{C} = \dfrac{5832}{2000} = 2.916^\circ\text{C}$

Rounding to one decimal place, we get $2.9^\circ\text{C}$.