Physics- heat of fusion?

Question

Determine the final temperature Tf, that results when 148 grams of ice at 0 °C are mixed with 300 grams of liquid water at 50 °C.

Specific heat of water: c = 1.00 cal/(gram*°C).

Heat of fusion for the ice - liquid water transition: cF = 79.7 cal/gram.

Answer 1

148x79.7 = 11795.6 cal

11795.6/300 = 39.3187 °C

When all the ice has melted the 300 g of water is at 50 - 39.3187 = 10.681 °C.

300*10.681/(300+148) = 7.15 °C

Answer 2

when the ice melts it removes 148 * 79.7 calories from the container. This chills the 300g by 148*79.7/300 degrees from 50 to Tchill.

You now have 148g at 0 and 300g at Tchill  

The final mix 448 * Tfinal = 300 * Tchill

I hope that's right!