Calculate the final temperature when 200 g of ice at 0°C is mixed with 500 g of water at 60°C.

$$m_{ice} = 200 \text{ g} = 0.2 \text{ kg}$$ $$m_{water} = 500 \text{ g} = 0.5 \text{ kg}$$

The formula for latent heat of fusion is $Q = mL_f$.

$$Q_{melt} = m_{ice} \times L_f$$ $$Q_{melt} = 0.2 \text{ kg} \times 336,000 \text{ J/kg}$$ $$Q_{melt} = 67,200 \text{ J}$$

Assuming no heat loss, the heat lost by the hot water equals the heat gained by the ice (to melt) plus the heat gained by the melted ice (to raise its temperature to the final temperature, $T_f$).
Heat lost by hot water = Heat gained by ice (melting) + Heat gained by melted ice (warming up)

$$m_{water} c_w (T_{initial\_water} - T_f) = Q_{melt} + m_{ice} c_w (T_f - T_{initial\_ice})$$

Substitute the known values:

$$0.5 \text{ kg} \times 4,200 \text{ J/kg}^\circ\text{C} \times (60^\circ\text{C} - T_f) = 67,200 \text{ J} + 0.2 \text{ kg} \times 4,200 \text{ J/kg}^\circ\text{C} \times (T_f - 0^\circ\text{C})$$

$$2,100 (60 - T_f) = 67,200 + 840 T_f$$ $$126,000 - 2,100 T_f = 67,200 + 840 T_f$$

Rearrange the terms to solve for $T_f$:

$$126,000 - 67,200 = 840 T_f + 2,100 T_f$$ $$58,800 = 2,940 T_f$$ $$T_f = \frac{58,800}{2,940}$$ $$T_f = 20^\circ\text{C}$$

The final temperature of the mixture is $\text{20}^\circ\text{C}$.