A

Step 1: Understand the situation
We have a block of metal at a high temperature (150ºC) and a calorimeter with water at a lower temperature (27ºC). When the metal block is dropped into the water, it will lose heat, while the water and calorimeter will gain heat until they reach a common final temperature (40ºC).

Step 2: Apply the principle of conservation of energy
The heat lost by the metal block will equal the heat gained by the water and the calorimeter. This can be expressed mathematically as:

$$m_{metal} c_{metal} (T_{initial, metal} - T_{final}) = (m_{water} + m_{cal}) c_{water} (T_{final} - T_{initial, water})$$

Where:

• $m_{metal}$ = mass of the metal = 0.20 kg
• $c_{metal}$ = specific heat of the metal (what we are trying to find)
• $T_{initial, metal}$ = initial temperature of the metal = 150ºC
• $T_{final}$ = final temperature = 40ºC
• $m_{water}$ = mass of water = volume × density = 150 cm³ × 1 g/cm³ = 0.15 kg
• $m_{cal}$ = water equivalent of the calorimeter = 0.025 kg
• $c_{water}$ = specific heat of water = 4184 J/(kg·ºC)

Step 3: Substitute values and solve
Now, plug the values into the equation:

$$0.20 imes c_{metal} imes (150 - 40) = (0.15 + 0.025) imes 4184 imes (40 - 27)$$

Now simplifying, we get:
$$0.20 imes c_{metal} imes 110 = 0.175 imes 4184 imes 13$$

Calculate the right side:
$$0.175 imes 4184 imes 13 = 0.175 imes 54412 = 9503.1$$
Thus, we have:
$$0.20 imes c_{metal} imes 110 = 9503.1$$

Divide both sides by (0.20 * 110):

$$c_{metal} = \frac{9503.1}{0.20 imes 110}$$
Calculating this gives:
$$c_{metal} = \frac{9503.1}{22} = 432.4 \text{ J/(kg·ºC)}$$

Final Result:
The specific heat of the metal is approximately 432.4 J/(kg·ºC). Therefore, option A is the correct answer.