The specific heat of water = 4200 J kg$$^{-1}$$ K$$^{-1}$$ and the latent heat of ice = $$3.4 \times 10^5$$ J kg$$^{-1}$$. 100 grams of ice at 0°C is placed in 200 g of water at 25°C. The amount of ice that will melt as the temperature of water reaches 0°C is close to (in grams)

We start by noting that the only significant energy exchange is between the warm water and the ice; the container and surroundings are assumed perfectly insulating. Because the ice is already at its melting point (0 °C) and the water finally cools to 0 °C, no part of the ice-water mixture ever rises above 0 °C. Hence the warm water will simply give up some of its thermal energy, cooling from 25 °C down to 0 °C, and that energy will go exclusively into melting a certain mass of ice.

First we calculate the heat lost by the water as it cools. The specific heat capacity formula is stated as

$$Q = m\,c\,\Delta T,$$

where $$m$$ is the mass of the substance, $$c$$ its specific heat capacity, and $$\Delta T$$ the change in temperature.

The warm water has

$$m_w = 200 \text{ g} = 0.200 \text{ kg},$$

$$c_w = 4200 \text{ J\,kg}^{-1}\text{K}^{-1},$$

$$\Delta T_w = 25^\circ\text{C} - 0^\circ\text{C} = 25 \text{ K}.$$

Substituting these values, we obtain

$$Q_{\text{lost}} = (0.200)\,(4200)\,(25) \text{ J}.$$

Now we multiply step by step:

$$4200 \times 25 = 105,000,$$

$$0.200 \times 105,000 = 21,000.$$

So,

$$Q_{\text{lost}} = 21,000 \text{ J}.$$

This 21,000 J of energy is absorbed by a portion of the ice, melting it while itself staying at 0 °C. For melting, the relevant formula is the latent heat relation

$$Q = m L,$$

where $$m$$ is the mass melted and $$L$$ is the latent heat of fusion of ice.

Given

$$L = 3.4 \times 10^{5} \text{ J\,kg}^{-1},$$

we set

$$Q_{\text{lost}} = Q_{\text{gained}},$$

$$21,000 = m\,L.$$

Solving for $$m$$ we have

$$m = \frac{21,000}{3.4 \times 10^{5}} \text{ kg}.$$

Carrying out the division step by step, we write

$$\frac{21,000}{340,000} = 0.0617647\ldots \text{ kg}.$$

To convert this to grams we multiply by 1000:

$$0.0617647\ldots \text{ kg} \times 1000 = 61.7647\ldots \text{ g}.$$

Rounded sensibly to three significant figures, the mass of ice that melts is approximately

$$\boxed{61.7 \text{ g}}.$$

Hence, the correct answer is Option A.