Answer :
We start by using the principle of conservation of energy. The heat gained by the water must equal the heat lost by the limestone. In other words,
$$
Q_{\text{water}} = Q_{\text{limestone}}.
$$
For the water, the energy gained is given by
$$
Q_{\text{water}} = m_{\text{w}}\, c_{\text{w}}\, (T_f - T_{w,i}),
$$
where
\( m_{\text{w}} = 75.0 \) g is the mass of the water,
\( c_{\text{w}} = 4.186 \) J/(g·°C) is its specific heat capacity,
\( T_{w,i} = 23.1^\circ \text{C} \) is its initial temperature, and
\( T_f = 51.9^\circ \text{C} \) is the final equilibrium temperature.
For the limestone, the heat lost is given by
$$
Q_{\text{limestone}} = m_{\text{l}}\, c_{\text{l}}\, (T_{l,i} - T_f),
$$
where
\( m_{\text{l}} = 62.6 \) g is the mass of the limestone,
\( c_{\text{l}} = 0.921 \) J/(g·°C) is its specific heat capacity, and
\( T_{l,i} \) is the unknown initial temperature of the limestone.
Setting the heat gained by the water equal to the heat lost by the limestone, we have:
$$
m_{\text{w}}\, c_{\text{w}}\, (T_f - T_{w,i}) = m_{\text{l}}\, c_{\text{l}}\, (T_{l,i} - T_f).
$$
To solve for \( T_{l,i} \), rearrange the equation:
$$
T_{l,i} = T_f + \frac{m_{\text{w}}\, c_{\text{w}}\, (T_f - T_{w,i})}{m_{\text{l}}\, c_{\text{l}}}.
$$
Now, substitute the known values:
1. Calculate the temperature change for the water:
$$
T_f - T_{w,i} = 51.9^\circ \text{C} - 23.1^\circ \text{C} = 28.8^\circ \text{C}.
$$
2. Compute the heat gained by the water:
$$
Q_{\text{water}} = 75.0 \times 4.186 \times 28.8 \approx 9041.76 \text{ J}.
$$
3. Calculate the product for the limestone:
$$
m_{\text{l}}\, c_{\text{l}} = 62.6 \times 0.921 \approx 57.6546 \text{ J/}^\circ\text{C}.
$$
4. Determine the temperature difference for the limestone:
$$
\frac{Q_{\text{water}}}{m_{\text{l}}\, c_{\text{l}}} \approx \frac{9041.76}{57.6546} \approx 156.83^\circ \text{C}.
$$
5. Finally, find the initial temperature of the limestone:
$$
T_{l,i} = 51.9^\circ \text{C} + 156.83^\circ \text{C} \approx 208.73^\circ \text{C}.
$$
Expressed to three significant figures, the initial temperature of the limestone is
$$
\boxed{209^\circ \text{C}}.
$$
$$
Q_{\text{water}} = Q_{\text{limestone}}.
$$
For the water, the energy gained is given by
$$
Q_{\text{water}} = m_{\text{w}}\, c_{\text{w}}\, (T_f - T_{w,i}),
$$
where
\( m_{\text{w}} = 75.0 \) g is the mass of the water,
\( c_{\text{w}} = 4.186 \) J/(g·°C) is its specific heat capacity,
\( T_{w,i} = 23.1^\circ \text{C} \) is its initial temperature, and
\( T_f = 51.9^\circ \text{C} \) is the final equilibrium temperature.
For the limestone, the heat lost is given by
$$
Q_{\text{limestone}} = m_{\text{l}}\, c_{\text{l}}\, (T_{l,i} - T_f),
$$
where
\( m_{\text{l}} = 62.6 \) g is the mass of the limestone,
\( c_{\text{l}} = 0.921 \) J/(g·°C) is its specific heat capacity, and
\( T_{l,i} \) is the unknown initial temperature of the limestone.
Setting the heat gained by the water equal to the heat lost by the limestone, we have:
$$
m_{\text{w}}\, c_{\text{w}}\, (T_f - T_{w,i}) = m_{\text{l}}\, c_{\text{l}}\, (T_{l,i} - T_f).
$$
To solve for \( T_{l,i} \), rearrange the equation:
$$
T_{l,i} = T_f + \frac{m_{\text{w}}\, c_{\text{w}}\, (T_f - T_{w,i})}{m_{\text{l}}\, c_{\text{l}}}.
$$
Now, substitute the known values:
1. Calculate the temperature change for the water:
$$
T_f - T_{w,i} = 51.9^\circ \text{C} - 23.1^\circ \text{C} = 28.8^\circ \text{C}.
$$
2. Compute the heat gained by the water:
$$
Q_{\text{water}} = 75.0 \times 4.186 \times 28.8 \approx 9041.76 \text{ J}.
$$
3. Calculate the product for the limestone:
$$
m_{\text{l}}\, c_{\text{l}} = 62.6 \times 0.921 \approx 57.6546 \text{ J/}^\circ\text{C}.
$$
4. Determine the temperature difference for the limestone:
$$
\frac{Q_{\text{water}}}{m_{\text{l}}\, c_{\text{l}}} \approx \frac{9041.76}{57.6546} \approx 156.83^\circ \text{C}.
$$
5. Finally, find the initial temperature of the limestone:
$$
T_{l,i} = 51.9^\circ \text{C} + 156.83^\circ \text{C} \approx 208.73^\circ \text{C}.
$$
Expressed to three significant figures, the initial temperature of the limestone is
$$
\boxed{209^\circ \text{C}}.
$$
