1. Given data

• Mass of aluminium sphere: $m_a = 0.20\,\text{kg}$
• Initial temperature of aluminium: $150^\circ \text{C}$
• Volume of water: $150\,\text{cc}$

Since $1\,\text{cc}$ of water has mass $1\,\text{g}$, $m_w = 150\,\text{g} = 0.150\,\text{kg}$

• Initial temperature of water: $27^\circ \text{C}$
• Water equivalent of calorimeter: $0.025\,\text{kg}$
• Final temperature: $40^\circ \text{C}$
• Specific heat of water: $1\,\text{cal/g}^\circ\text{C}$
• Conversion: $1\,\text{cal} = 4.2\,\text{J}$

2. Heat lost by aluminium = Heat gained by water + calorimeter

Let the specific heat of aluminium be $c$ in $\text{J/kg}^\circ\text{C}$.

To avoid unit confusion, first solve in calories.

Let specific heat of aluminium in $\text{cal/g}^\circ\text{C}$ be $s$. Then later convert to SI units.

Mass of aluminium: $0.20\,\text{kg} = 200\,\text{g}$

Temperature fall of aluminium: $150 - 40 = 110^\circ \text{C}$

So heat lost by aluminium is: $Q_a = 200\,s\,(110)$

3. Heat gained by water and calorimeter

Total water equivalent absorbing heat: $150 + 25 = 175\,\text{g}$

Temperature rise: $40 - 27 = 13^\circ \text{C}$

Hence heat gained: $Q_g = 175 \times 1 \times 13 = 2275\,\text{cal}$

4. Apply heat balance

$200\,s\,(110) = 2275$

$22000s = 2275$

$s = \frac{2275}{22000} = 0.1034\,\text{cal/g}^\circ\text{C}$

5. Convert to J/kg$^\circ$C

Since $1\,\text{cal/g}^\circ\text{C} = 4200\,\text{J/kg}^\circ\text{C}$

Therefore, $c = 0.1034 \times 4200$

$c \approx 434\,\text{J/kg}^\circ\text{C}$

6. Match with options

The value is $\boxed{434\,\text{J/kg}^\circ\text{C}}$

So the correct option is D.