Answer to Question #259329 in General Chemistry for Uomi Chihiro

Question

A 458 mL of water at 21.5 ⁰C is in a calorimeter. If a 758.7 g gold at 99.7 ⁰C is put into the calorimeter, the temperature becomes 25.3 ⁰C. Sp. Heat of Gold = 0.129 J/g-K

If the calorimeter’s heat capacity is 1.53 J/K, what will the final temperature be if the same gold with the same initial temperature is added?

Answer 1

Q=mc∆T

=458×(21.5-25.3)×1.53

= -2662.82

= 758.7(99.7-25.3)×0.129= 437575.814

= 437575.814/2662.82=164.33°C