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

User Elida
by
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1 Answer

12 votes

Answer:

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

User Jose Gomez
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4.5k points
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