Based on the thermodynamic properties provided for water, determine the amount of energy needed for 3.40 kg of water to go from -19.0 °C to 64.0 °C.
Melting point | 0.0 | °C |
Boiling point | 100.0 | °C |
?Hfus | 6.01 | kJ/mol |
?Hvap | 40.67 | kJ/mol |
cp (s) | 37.1 | J/mol ·°C |
cp (l) | 75.3 | J/mol ·°C |
cp (g) | 33.6 | J/mol ·°C |
3.40 kg of water = 3400 gm water = mass / molar mass = 3400 / 18 = 188.9 mole.
amount of energy needed = n1 * Cp(s) * dT1 (- 19 to o oC) + n1 * delta Hfus + n1 * Cp(l) * dT2 (from 0 to 64 oC)
amount of energy needed = 188.9 * 37.1 * (0 + 19) + 188.9 * (6.01 * 1000) + 188.9 * 75.3 * (64 - 0)
amount of energy needed = 133155.6 + 1135289 + 910346.9
amount of energy needed = 2178791.5 J = 2178.8 KJ (answer)
Based on the thermodynamic properties provided for water, determine the amount of energy needed for 3.40...
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Based on the thermodynamic properties provided for water, determine the amount of energy needed for 2.60 kg of water to go from -12.0°C to 81.0°C. Value Units 0.0 100.0 °C Property Melting point Boiling point A Hus A Hvap cp (s) 6.01 40.67 37.1 kJ/mol kJ/mol J/mol . °C J/mol.°C J/mol. °C 610 75.3 33.6
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Based on the thermodynamic properties provided for water, determine the amount of energy needed for 2.60 kg of water to go from -12.0°C to 81.0°C. Units °C Property Melting point Boiling point A Hius ΔΗ,ap Cp (s) Value 0.0 100.0 6.01 40.67 37.1 75.3 kJ/mol kJ/mol J/mol. °C J/mol - C J/mol. °C Co (8) 33.6
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Based on the thermodynamic properties provided for water,
determine the energy change when the temperature of 0.350 kg of
water decreased from 117 °C to 23.0 C.
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