m = mass of ethyl alchohol = 0.539 kg
Lv = heat of vaporization = 879 kJ/kg
c = specific heat = 2.43 J/(kg-K)
Lf = latent heat of fusion = 109 kJ/kg
T = change in temperature = 78 - (- 114) = 192
Q = Total Heat
Q = Heat of vaporization + Heat due to change in temperature + heat of fusion
Q = m Lv + m c T + m Lf
Q = (0.539) (879) + (0.539) (2.43) (192) + (0.539) (109)
Q = 784.01 kJ
Question 5 Ethyl alcohol has a boiling point of 78.0 °C, a freezing point of -114...
Ethyl alcohol has a boiling point of 78.0 ˚C, a freezing point of -114 ˚C, a heat of vaporization of 879 kJ/kg, a heat of fusion of 109 kJ/kg, and a specific heat of 2.43 kJ/kg·K. How much energy must be removed from 0.689 kg of ethyl alcohol that is initially a gas at 78.0 ˚C so that it becomes a solid at -114 ˚C?
Ethyl alcohol has a boiling point of 78.0 ˚C, a freezing point of -114 ˚C, a heat of vaporization of 879 kJ/kg, a heat of fusion of 109 kJ/kg, and a specific heat of 2.43 kJ/kg·K. How much energy must be removed from 0.683 kg of ethyl alcohol that is initially a gas at 78.0 ˚C so that it becomes a solid at -114 ˚C?
Ethyl alcohol has a boiling point of 78.0 degree C, a freezing point of -114 degree C, a heat of vaporization of 879 kJ/kg, a heat of fusion of 109 kJ/kg, and a specific heat of 2.43 kJ/kg-K. How much energy must be removed from 0.567 kg of ethyl alcohol that is initially a gas at 78.0 degree C so that it becomes a solid at -114 degree C?
Chapter 18, Problem 039 Ethyl alcohol has a boiling point of 78.0 °C, a freezing point of -114 °C, a heat of vaporization of 879 kJ/kg, a heat of fusion of 109 kJ/kg, and a specific heat of 2.43 kJ/kg K. How much energy must be removed from 0.765 kg of ethyl alcohol that is initially a gas at 78.0 °C so that it becomes a solid at -114 C? Number Units the tolerance is +/-2%
Ethyl alcohol has a boiling point of 78.0 degrees Celsius, a freezing point of -114 degrees Celsius, a latent heat of vaporization of 879 kJ/K, a latent heat of fusion of 109 kJ/K, and a specific heat of 2.43 kJ/kg.K. How much energy must be removed from 0.510 kg of ethyl alcohol that is initially a gas at 78.0 degrees Celsius so that it becomes a solid at -114 degrees Celsius?
Need help on those two problems! Question 4 Incorrect. Ethyl alcohol has a bolling point of 780°C a freezing point of-1 14 С a heat of vaporization of 879 kJkg a heat of fusion of 109 kle and a specific C? Number778 heat of 243kJkgK. Howmuch enersy must be removed from Os535 kg of ethylalcoho that sinitialy gas at 780 C so that i becomes a sold a 114 Units kg eTextbook and Media Hint Attempts: 1 of 6 used...
What is the freezing point of an alcohol in °C to the nearest degree if its heat of fusion(AHlusº) +9.37 kJ/mol and its entropy of fusion ASfusº = 50.6 J/K.mol.
potassium metal has melting point of 63.65 C and a boiling point of 774 C Enthalpy fusion = 14.37 cal/g enthaply of vaporization = 490.3 cal/g Specific heat @ solid = .1759 cal/gC specfic heat @ liquid =.5425 cal /gC specfic heat @ gas = .2629 cal /gC a) draw heatung curve that us heated from 23C to 1200c b) Calculate total heat required in units of kJ for 39.5 g of potassium
Pure ethanol has a boiling point of 78.0 °C. What is the boiling point of a solution that contains 12.7 grams of KI (MW = 166 g/mol) dissolved in 125 grams of ethanol (MW = 46.0 g/mol)? Assume that KI is a strong electrolyte in ethanol. (Ethanol: kb= 1.22 °C/m)
Molal Boiling-Point-Elevation and Freezing-Point-Depression Solvent Normal Boiling Point (∘C) Kb (∘C/m) Normal Freezing Point (∘C) Kf (∘C/m) Water, H2O 100.0 0.51 0.0 1.86 Benzene, C6H6 80.1 2.53 5.5 5.12 Ethanol, C2H5OH 78.4 1.22 -114.6 1.99 Carbon tetrachloride, CCl4 76.8 5.02 -22.3 29.8 Chloroform, CHCl3 61.2 3.63 -63.5 4.68 Part E freezing point of 2.02 g KBr and 4.84 g glucose (C6H12O6) in 187 g of water Part F boiling point of 2.02 g KBr and 4.84 g glucose (C6H12O6) in...