Question
A 1000 g sample of metal with a specific heat of 0.50 J/goC is heated to 100.0oC and then placed in a 50.0 g sample of water at 20.0oC. (Hint: Specific heat of water at 20.0oC is 4.182 J/goC). What is the final temperature of the metal and the water? Write your answer to the correct number of significant figures in scientific notation.
Answers
heat gained by water equals heat lost by metal
50.0 * (t - 20.0) * 4.182 =
... 1000 * (100.0 - t) * 0.50
50.0 * (t - 20.0) * 4.182 =
... 1000 * (100.0 - t) * 0.50
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