A

Step 1: Let's denote the mass of water and ice as 'm'. The specific heat capacity of water is approximately c_w = 4.18 J/g°C, and the specific heat capacity of ice is approximately c_i = 2.09 J/g°C.
Step 2: The heat lost by the water as it cools down to the final temperature T_f can be expressed as:
$$ Q_{lost} = m imes c_w imes (T_i - T_f) = m imes 4.18 imes (50 - T_f) $$
Step 3: The heat gained by the ice as it warms up to 0°C and then melts and further warms to the final temperature T_f can be expressed as:
$$ Q_{gained} = m imes c_i imes (0 - (-40)) + m imes L_f + m imes c_w imes (T_f - 0) $$
where L_f = 334 J/g is the latent heat of fusion of ice. Thus:
$$ Q_{gained} = m imes 2.09 imes (40) + m imes 334 + m imes 4.18 imes T_f $$
Step 4: Setting heat lost equal to heat gained:
$$ m imes 4.18 imes (50 - T_f) = m imes 2.09 imes (40) + m imes 334 + m imes 4.18 imes T_f $$
Cancelling 'm' from both sides (since they're equal), we get:
$$ 4.18(50 - T_f) = 2.09(40) + 334 + 4.18T_f $$
Step 5: Simplifying further:
$$ 209 - 4.18T_f = 83.6 + 334 + 4.18T_f $$
$$ 209 - 83.6 - 334 = 4.18T_f + 4.18T_f $$
$$ -208.6 = 8.36T_f $$
Step 6: Solving for T_f:
$$ T_f = \frac{-208.6}{8.36} \approx -24.98°C $$
So, rounding, we find a final temperature of approximately -25°C.
Therefore, the final temperature of the mixture is: -25°C, which corresponds to option A.