Lab 11 - = 58.1 C ̊ Q sub hot = cm sub hot T sub hot – T sub final =(4.187 kJ/kgK × 200mL ×(88.9 C-58.1 C ̊ ̊ =(837.4 kJ/kgK/mL ×(30.8 C ̊

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Lab 11 Calorimetry and Latent Heat 1. 150mL of hot water T sub h = 85.7 C ̊ 150mL of cold water T sub c = 27.6 C ̊ T sub f, measured = 57.5 C ̊ T sub f, calculated = T sub h + T sub c / 2 = 85.7 C + 27.6 C / 2 ̊ ̊ = 113. 3 C / 2 ̊ = 56.65 C ̊ Q sub hot = cm sub hot (T sub hot – T sub final) = (4.187 kJ/kgK × 150mL) × (85.7 C – 56.65 C) ̊ ̊ = (628.05 kJ/kgK/mL) × (29.05 C) ̊ = 18244.8525 J Q sub cold = cm sub cold (T sub final – T sub cold) = (4.187kJ/kgK × 150mL) × (56.65 C – 27.6 C) ̊ ̊ = (628.05kJ/kgK/mL) × (29.05 C) ̊ = 18244.8525 J 2. 200mL of cold water T sub cold = 27.3 C ̊ 200mL of hot water T sub hot =88.9 C ̊ T sub f, measured = 44.9 C ̊ T sub f, calculated = T sub c + T sub h / 2 = 27.3 C + 88.9 C / 2 ̊ ̊ = 116.2 C /2 ̊
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Unformatted text preview: = 58.1 C ̊ Q sub hot = cm sub hot ( T sub hot – T sub final) = (4.187 kJ/kgK × 200mL) × (88.9 C -58.1 C) ̊ ̊ = (837.4 kJ/kgK/mL) × (30.8 C) ̊ = 25791.92 J Q sub cold =cm sub cold ( T sub final – T sub cold) = (4.187kJ/kgK × 200mL) × (58.1 C – 27.3 C) ̊ ̊ = (837.4kJ/kgK/mL) × (30.8 C) ̊ = 25791.92 J 3. 200 mL of cold water Time, min T C ̊ Initial 27.2 1 min 37.6 2 min 52.0 3 min 62.4 4 min 71.0 5 min 80.0 6 min 91.2 7 min 97.2 8 min 100.5 Mass of beaker = 229.6 g Mass of beaker + 200mL of water =431.2 g Weight of beaker after = 405.9 g L /c = L = ? Q = m L L = Q /m = 25791.92 J / 200 mL = 128. 9596 J/mL L/c = (128.9596J/mL) / (4.187 kJ/kgK) = 30.8kgK/KmL...
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This note was uploaded on 10/12/2011 for the course CORC 1331 taught by Professor Staff during the Fall '10 term at CUNY Brooklyn.

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Lab 11 - = 58.1 C ̊ Q sub hot = cm sub hot T sub hot – T sub final =(4.187 kJ/kgK × 200mL ×(88.9 C-58.1 C ̊ ̊ =(837.4 kJ/kgK/mL ×(30.8 C ̊

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