Ice at $$-20°C$$ is added to 50 g of water at $$40°C$$. When the temperature of the mixture reaches $$0°C$$, it is found that 20 g of ice is still unmelted. The amount of ice added to the water was close to (Specific heat of water = 4.2 J/g/°C, Specific heat of Ice = 2.1 J/g/°C, Heat of fusion of water at $$0°C$$ = 334 J/g)

We have 50 g of liquid water whose initial temperature is $$40^{\circ}\text{C}$$. Some mass of ice, say $$m\; \text{g}$$, is added at $$-20^{\circ}\text{C}$$. After thermal equilibrium is attained, the common temperature becomes $$0^{\circ}\text{C}$$ and 20 g of ice still remains unmelted. Therefore only $$m-20\;\text{g}$$ of the originally added ice has actually melted.

The process involves heat exchange between two parts:

1. Cooling of the warm water from $$40^{\circ}\text{C}$$ down to $$0^{\circ}\text{C}$$.
   Using the formula for sensible heat $$Q = mc\Delta T$$, the heat released is

$$ Q_{\text{water}} = (50\;\text{g})(4.2\;\text{J g}^{-1}\!^{\circ}\text{C}^{-1})(40^{\circ}\text{C}) = 50 \times 4.2 \times 40 = 8400\;\text{J}. $$

2. Warming and partial melting of the ice added at $$-20^{\circ}\text{C}$$.

   (a) First, the ice must be warmed from $$-20^{\circ}\text{C}$$ to $$0^{\circ}\text{C}$$.
     Again with $$Q = mc\Delta T$$,

$$ Q_{\text{warm}} = m \,(2.1\;\text{J g}^{-1}\!^{\circ}\text{C}^{-1})(20^{\circ}\text{C}) = 42\,m\;\text{J}. $$

   (b) Next, only $$m-20\;\text{g}$$ of this ice actually melts at $$0^{\circ}\text{C}$$. The latent heat needed is given by

$$ Q_{\text{melt}} = (m-20)\;\text{g}\times 334\;\text{J g}^{-1} = 334\,(m-20)\;\text{J}. $$

The total heat absorbed by the ice is therefore

$$ Q_{\text{ice}} = Q_{\text{warm}} + Q_{\text{melt}} = 42\,m + 334\,(m-20) = 42\,m + 334\,m - 6680 = 376\,m - 6680\;\text{J}. $$

Because no heat is lost to the surroundings, the heat released by the water must equal the heat gained by the ice:

$$ Q_{\text{water}} = Q_{\text{ice}}. $$

Substituting the expressions we have just found,

$$ 8400 = 376\,m - 6680. $$

Now we solve algebraically for $$m$$. First add 6680 J to both sides:

$$ 8400 + 6680 = 376\,m, $$ $$ 15080 = 376\,m. $$

Dividing both sides by 376,

$$ m = \frac{15080}{376} \approx 40.1\;\text{g}. $$

This mass is closest to 40 g among the given choices.

Hence, the correct answer is Option D.