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# 14-3p597 - 14-3p597.wpd Page 1 of 2 Summary P597#8-11...

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14-3p597.wpd Page 1 of 2 Summary: P597 #8-11; P599 #12-17; P600 #1-7 P597 8) C(ethanol) = 2.46 J/g C 0 Q = mC Ä T = 100 g X 2.46 J/g C X (50 C - 25 C) = 6.15 X 10 J 0 0 0 3 9) C(water(l)) = 4.184 J/g C 0 Q = mC Ä T = 120.0g X 4.184 J/g C X 9.1 C = 4.6 X 10 J 0 0 3 10) The second beaker gains more thermal energy because a larger mass of an identical substance is increased to the same temperature difference. 11) a) Beaker 1: Q = + 4.22 kJ Beaker 2: Q = + 4.6 kJ b) Q = + 487 kJ c) Q = -290 kJ P599 12) a) m = Q ÷ C Ä T b) C = Q ÷ m Ä T c) Ä T = Q ÷ mC 13) i f a) T = T - Q ÷ mC f i b) T = T + Q ÷ mC 3(l) 14) C(NH ) = 4.70 J/g C 0 Q = mC Ä T = 789 g X 4.70 J/g C X (82.7 C - 25.0 C) = 214 kJ 0 0 0 15) C = Q ÷ m Ä T C = -4937.50 J ÷ (250.00 g X -25.00 C) = 0.7900 J/g C 0 0 The substance is granite. 16) from the heated water: Q = mC Ä T Q = 75.0 g X 4.184 J/g C X 8.5 C = 2.7 kJ 0 0 from the cooled down metal:

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14-3p597.wpd Page 2 of 2 C = Q ÷ m Ä T C = 2.7 X 10 J ÷ (250.0 g X (98.0 C - 28.5 C) = 0.16 J/g C 3 0 0 0 17) m w Q = -Q (m is metal, w is water) Q = mC Ä T therefore: m m f im w w f iw m C (T -T ) = -m C (T - T ) m m f w w f w w iw m m im m C T + m C T = m C T + m C T f w w iw m m im m m w w T = (m C T + m C T )/(m C + m C ) f T = (192X4.184X15.0 + 45.5X0.129X80.5) / (45.5 X 0.129 + 192X4.184) f T = 12552 / 809.2 = 15.5 C
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