How much heat in kJ gets rid of condensing 7.13 g of water vapor at 100 ° C and then cools the water formed to 25 ° C? [ΔHvap = 40.7 kJ / mol at 100 ° C and the natural heat of water c (H2O (l)) = 4.18 J g-1 ° C-1] Note. look up the units in the calculations and give the answer in kJ
We have 7.13 g of water vapor at 100 ° C So, moles of water present = 7.13 g / (18 g/mol) = 0.396 moles Condensation of water vapour to water at 25°C has two steps : (i) water vapour to water at 100 C heat lost is latent heat of vapourization , we have ΔHvap = 40.7kJ/mol So, amount of heat lost during this step , Q1 = 0.396 moles*40.7kJ/mol = 16.12 kJ (ii) It cools to 25 ° C Heat is lost during , Q2 = m*ΔT* c ( m = mass of water ;ΔT = temp. change (in degree) : c = specific/natural heat of water = 4.18 J g-1°C-1 Q2 = 7.13 g* 75 oC * 4.18 J g-1°C-1 = 2235.2 J = 2.24 kJ So, total amount of heat gets rid = Q1 + Q2 = 18.36 kJ
How much heat in kJ gets rid of condensing 7.13 g of water vapor at 100...
how much heat is released when 10.0 g of steam (water vapor ) at 105.0 C is cooled to liquid water at 25 C? S(water) = 4.18 J/g.C. ... S(steam) = 2.01 j/ g.C the heat of fusion of water is 6.02 KJ/ mol. The heat of vaporization of water is 40.7 KJ/mol
How much energy is required to heat 36.0 g H2O from a liquid at 55.0°C to a gas at 150.0°C? The following physical data may be useful. Molar Mass(H2O) = 18.0 g/mol ΔHvap = 40.7 kJ/mol Cs;liquid = 4.18 J/g oC Cs;gas = 2.01 J/goC Tb(H2O) = 100.0 oC
How much energy is required to heat 36.0 g H2O from a liquid at 65°C to a gas at 115°C? The following physical data may be useful. ΔHvap = 40.7 kJ/mol Cliq = 4.18 J/g°/sup>C Cgas = 2.01 J/g°/sup>C Csol = 2.09 J/g°/sup>C Tmelting = 0°/sup>C Tboiling = 100°/sup>C
Using the provided data, calculate the amount of heat, in kJ, required to warm 21.7 g of solid water, initially at -10. °C, to gaseous water at 112. °C. water molar mass 18.0153 g/mol melting point 0. °C boiling point 100. °C ΔHfus 6.02 kJ/mol ΔHvap at bp 40.7 kJ/mol Cs, solid 2.09 J/g⋅°C Cs, liquid 4.18 J/g⋅°C Cs, gas 1.87 J/g⋅°C 72.4 kJ 57.6 kJ 58.5 kJ 66.3 kJ 13.9 kJ
Using the provided data, calculate the amount of heat, in kJ, that must be removed to cool 22.9 g of gaseous water, initially at 122. °C, to solid water at -29. °C. water molar mass 18.0153 g/mol melting point 0. °C boiling point 100. °C ΔHfus 6.02 kJ/mol ΔHvap at bp 40.7 kJ/mol Cs, solid 2.09 J/g⋅°C Cs, liquid 4.18 J/g⋅°C Cs, gas 1.87 J/g⋅°C
Using the provided data, calculate the amount of heat, in kJ, that must be removed to cool 18.8 g of gaseous water, initially at 110. °C, to solid water at -12. °C. water molar mass 18.0153 g/mol melting point 0. °C boiling point 100. °C ΔHfus 6.02 kJ/mol ΔHvap at bp 40.7 kJ/mol Cs, solid 2.09 J/g⋅°C Cs, liquid 4.18 J/g⋅°C Cs, gas 1.87 J/g⋅°C 40.8 kJ 67.7 kJ 37.9 kJ 57.4 kJ 16.3 kJ
How much energy (heat) is required to convert 52.0 g of ice at -10.0 C to steam at 100 C?Specific heat of ice: 2.09 J/g * C DHfus = 6.02 kJ/molSpecific heat of water: 4.18 J/g * C DHvap = 40.7 kJ/molSpecific heat of steam: 1.84 J/g * C
Part A Calculate the enthalpy change, ΔH, for the process in which 42.4 g of water is converted from liquid at 15.3 ∘C to vapor at 25.0 ∘C . For water, ΔHvap = 44.0 kJ/mol at 25.0 ∘C and Cs = 4.18 J/(g⋅∘C) for H2O(l). How many grams of ice at -16.2 ∘C can be completely converted to liquid at 25.5 ∘C if the available heat for this process is 4.77×103 kJ ? For ice, use a specific heat of...
Calculate the amount of heat that must be absorbed by 100.0 g of water at 20.0°C to convert it to steam (water vapor) at 110.0°C. Given: Specific heats: (liq) = 4.18 J/g·°C (steam) = 1.84 J/g·°C DHvap = 40.7 kJ/mol
How much heat (in kJ) is required to convert 431 g of liquid H2O at 23.6°C into steam at 148°C? (Assume that the specific heat of liquid water is 4.184 J/g·°C, the specific heat of steam is 2.078 J/g·°C, and that both values are constant over the given temperature ranges. The normal boiling point of H2O is 100.0°C. The heat of vaporization (ΔHvap) is 40.65 kJ/mol.)