Calculate the specific heat capacities of Zinc and Brass using the principle of calorimetry and the provided data.

Heat loss by zinc = Heat gained by water

$$m_Z c_Z \Delta T_Z = m_W c_W \Delta T_W$$

Given:
$m_W = 0.303 \, \text{kg}$
$c_W = 4180 \, \text{J/kg K}$
$\Delta T_W = 1 \, \text{°C} = 1 \, \text{K}$
$m_Z = 0.08 \, \text{kg}$
$\Delta T_Z = 73 \, \text{K}$

$$c_Z = \frac{(0.303 \, \text{kg})(4180 \, \text{J/kg K})(1 \, \text{K})}{(0.08 \, \text{kg})(73 \, \text{K})}$$

$$c_Z = \frac{1266.54 \, \text{J}}{5.84 \, \text{kg K}}$$ $$c_Z = 216.8732... \, \text{J/kg K}$$

Rounding to two decimal places:

$$c_Z = \boxed{\text{216.87 J/kg K}}$$

Brass

Heat loss by brass = Heat gained by water

$$m_B c_B \Delta T_B = m_W c_W \Delta T_W$$

Given:
$m_W = 0.277 \, \text{kg}$
$c_W = 4180 \, \text{J/kg K}$
$\Delta T_W = 1.5 \, \text{°C} = 1.5 \, \text{K}$
$m_B = 0.08 \, \text{kg}$
$\Delta T_B = 72.5 \, \text{K}$

$$c_B = \frac{(0.277 \, \text{kg})(4180 \, \text{J/kg K})(1.5 \, \text{K})}{(0.08 \, \text{kg})(72.5 \, \text{K})}$$

$$c_B = \frac{1737.69 \, \text{J}}{5.8 \, \text{kg K}}$$ $$c_B = 299.6017... \, \text{J/kg K}$$

Rounding to two decimal places:

$$c_B = \boxed{\text{299.60 J/kg K}}$$