Question

how much heat is necessary to change 30.0g of water at 25.0c to steam at 120.0c?

Answer 1

Initially water is at 25 C Water is first heated to 100 C from 25oC in liquid phase. Heat is then required to change the phase from liquid to vapor state and then heat is required to change the temperature from 100 to 120 C in vapor ph Mass of water (m) = 30 g Molar mass of water 18.02 g/mol Moles of water mass/ molar mass 30g/ (18.02 g/mol) 1.66 mol Heat required to change the temperature from 25 to 100 C in liquid phase is: qi m*Specific heat of water*change in temperature 30g *4.18J/g. C (100-25) C 9405 J 9.405 ase. kJ Heat required to to change the phase from liquid to XARr state is: q, = moles of water * ▲Han-1.66 mol *40.65 kJ /mol = 67.675 kJ heat required to change the temperature from 100 to 120.C in yapor phase is: q, = m * specific heat of vapor" change in temperature-30 g *1.841/g°C" (120-100)«C = 1104 J = 1.104 kJ Total heat required qtq 9.405 67.675 1.104 78.18 kJ