A 332.3 gram block of solid water is heated from -77.4°C to 51.7°C. Determine the total energy (kJ) absorbed during this process.
M.P. 0°C
B.P. 100°C
ΔHf = 6.02 kJ/mole
ΔHv = 40.6 kJ/mol
cp(g) = 1.97J/g°C
cp(l) = 4.18J/g°C
cp(s) = 1.85 J/g°C
A 332.3 gram block of solid water is heated from -77.4°C to 51.7°C. Determine the total energy.....?
2015-04-16 1:02 am
回答 (1)
2015-04-16 4:18 pm
(1.85 J/g°C) x (332.3 g) x (0 - (-77.4)) °C = 47582 J to warm the ice to its melting point
(6.02 kJ/mol) x ((332.3 g H2O) / (18.01532 g H2O/mol)) = 111.041 kJ = 111041 J to melt the ice
(4.18J/g°C) x (332.3 g) x (51.7 - 0) °C = 71812 J to warm the melted ice to 51.7 °C
47582 J + 111041 J + 71812 J = 230435 J = 230 kJ total
(6.02 kJ/mol) x ((332.3 g H2O) / (18.01532 g H2O/mol)) = 111.041 kJ = 111041 J to melt the ice
(4.18J/g°C) x (332.3 g) x (51.7 - 0) °C = 71812 J to warm the melted ice to 51.7 °C
47582 J + 111041 J + 71812 J = 230435 J = 230 kJ total
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