In SI units: kJ/kmol
Molar heat of fusion: the heat (enthalpy, energy) needed to transform a solid in liquid (expressed in kJ/mol). Molar heat of vaporization: the heat (enthalpy, energy) needed to transform a liquid in gas (expressed in kJ/mol).
The quantity of heat required to change the temperature of 1g of a substance by 1 celsius is defined as its specific heat or specific heat capacity.Translating the question into "math-speak" will give you the units: it wants to know heat per gram per degree celsius.Heat = J (or cal), per means divide, gram = g, degree celsius = oC, soJ/(g)(oC), which is the unit for specific heat capacity!
Fusion refers to the phase change from liquid to solid (or vv) and vaporization refers to the phase change from liquid to gas (or vv). Heat is either absorbed or released from the substance when these events occur. We use the molar enthalpies to calculat exactly how much heat is transferred during these processes. Ex .. If we are to boil (vaporize) 3 moles of some substance with a known molar enthalpy of vaporization of 120 KJ/mole then we multiply. 3 moles x 120 KJ/mole = 360 KJ of energy is needed for the vaporization to take place.
Molar heat capacity is the amount of heat required to raise the temperature of one mole of a substance by one degree Celsius (or one Kelvin). It is an important thermodynamic property that reflects how a substance absorbs heat. The units of molar heat capacity are typically expressed in joules per mole per degree Celsius (J/mol·°C) or joules per mole per Kelvin (J/mol·K).
thermol
It is a known fact : Molar heat of sublimation = molar heat of fusion + molar heat of vaporization so, molar heat of vaporization = molar heat of sublimation - molar heat of fusion Mv = 62.3 kJ/mol - 15.3 kJ/mol Mv = 47 kJ/mol.
The molar heat of vaporization can be estimated by using the Clausius-Clapeyron equation, which relates the vapor pressure of a substance to its temperature and molar heat of vaporization. By knowing the temperature change and the corresponding increase in vapor pressure, calculations can be made to determine the molar heat of vaporization.
Molar heat of fusion: the heat (enthalpy, energy) needed to transform a solid in liquid (expressed in kJ/mol). Molar heat of vaporization: the heat (enthalpy, energy) needed to transform a liquid in gas (expressed in kJ/mol).
The heat absorbed during vaporization is called the heat of vaporization. For carbon tetrachloride, the heat of vaporization is 30.5 kJ/mol. To calculate the heat absorbed when 75 g of CCl4 vaporizes, you would first convert grams to moles using the molar mass of CCl4. Then, use the heat of vaporization to calculate the total heat absorbed.
The latent heat of evaporation
The quantity of heat required to change the temperature of 1g of a substance by 1 celsius is defined as its specific heat or specific heat capacity.Translating the question into "math-speak" will give you the units: it wants to know heat per gram per degree celsius.Heat = J (or cal), per means divide, gram = g, degree celsius = oC, soJ/(g)(oC), which is the unit for specific heat capacity!
Fusion refers to the phase change from liquid to solid (or vv) and vaporization refers to the phase change from liquid to gas (or vv). Heat is either absorbed or released from the substance when these events occur. We use the molar enthalpies to calculat exactly how much heat is transferred during these processes. Ex .. If we are to boil (vaporize) 3 moles of some substance with a known molar enthalpy of vaporization of 120 KJ/mole then we multiply. 3 moles x 120 KJ/mole = 360 KJ of energy is needed for the vaporization to take place.
First, calculate the number of moles of C2H5Cl in 0.38 g. Then, use the molar heat of vaporization to find the heat absorbed for this number of moles. Finally, convert the heat from per mole to kilojoules. The heat absorbed when 0.38 g of chloroethane vaporizes at its boiling point will be 3.58 kJ.
Latent Heat of Evaporation, or Evaporation Enthalpy. It is given in units of energy over unit of mass, i.e., KJ/Kg.
The heat of vaporization of water is 40.79 kJ/mol. First, determine the number of moles in 24.40 grams of water. Then, convert moles to joules using the molar heat of vaporization. This will give you the amount of heat needed to vaporize 24.40 grams of water.
Vaporization heat of water: 40.65 kJ/mol or 2257 kJ/kg or 539.423 calories per gram (very outdated units!)
First things first: it's actually spelled "enthalpy", which might be why you're not finding it.If you want a number, you will need to specify a substance.If you just want to know what it means, then in simple terms it's the amount of energy required to evaporate one mole of the substance.