1. Given data

• Mass of ice: $m_i = 100\text{ g} = 0.1\text{ kg}$ at $0^\circ\text{C}$
• Mass of water: $m_w = 200\text{ g} = 0.2\text{ kg}$ at $25^\circ\text{C}$
• Specific heat of water: $c = 4200\,\text{J kg}^{-1}\text{K}^{-1}$
• Latent heat of fusion of ice: $L = 3.4 \times 10^5\,\text{J kg}^{-1}$

We need the amount of ice melted when the final temperature becomes $0^\circ\text{C}$.

2. Heat lost by water in cooling from $25^\circ\text{C}$ to $0^\circ\text{C}$

$Q = m_w c \Delta T$

$Q = 0.2 \times 4200 \times 25$

$Q = 21000\,\text{J}$

3. This heat is used to melt some ice at $0^\circ\text{C}$

If mass melted is $m$, then

$Q = mL$

So,

$m = \frac{Q}{L} = \frac{21000}{3.4 \times 10^5}$

$m = 0.06176\,\text{kg}$

Converting to grams:

$m = 61.76\,\text{g}$

4. Closest option

$m \approx 61.7\,\text{g}$

So the correct option is B.

5. Comparison with stored answer

Stored correct answer: B

Our derived answer: B

They agree.