Question

what is the final temperature when 123.55 grams of ice is placed at 50 degrees celsius of 304.79 grams of water?

Answer 1

Below in bold.

Step-by-step explanation:

The ice is at 273 degrees K and the water at 323 degrees K.

123.55* 273 + 304.79*323 = (123.55+304.79)* t

t =(123.55* 273 + 304.79*323) / (123.55+304.79)

=308.58 degrees K

= 35.6 degrees C