Lecture 10 notes (exam 2 review)

Waterboilsh2ol h2ogendothermic exothermic

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Unformatted text preview: is: Added to a system 0.902 0.720 0.451 0.385 0.129 24.3 8.65 25.1 24.4 25.4 4.70 4.184 2.46 2.42 2.06 0.861 80.1 75.3 112. 149. 37.1 132. 72.3 1.76 0.84 0.79 Heat Capacity Example 24.1 kJ of energy is lost by a 250. g aluminum block. If the block is initially at 125.0°C what will be its final temperature? (cAl = 0.902 J g‐1 °C‐1) q = m c ΔT ΔT = q / (m c) heat is lost, q is negative ΔT = −24.1 x 103 J /(250. g x 0.902 J g-1 C-1) ΔT = Tfinal – Tinital = −107 C Thus Tfinal = ΔT + Tinital = −107 + 125°C = 18°C Freezing and Melting (Fusion) During freezing (or melting) Removed from a system q is negative the change is exothermic. Water Boils: H2O(l) → H2O(g) endothermic exothermic Work occurs as the sample expands or contracts. Overall: ΔE = q + w • Substance loses (or gains) E, but… ...
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This note was uploaded on 03/02/2014 for the course CHEM 101 taught by Professor Multipleprofs during the Spring '07 term at Drexel.

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