What happens when we add energy to a solid at constant pressure gas

What are the energy changes that occur?  H =  H fus x # mols  = m x C solid x  t  = m x C liquid x  t  H =  H vap x # mols  = m x C gas x  t

What would these changes look like on a temperature versus energy graph?  H =  H fus x # mols  = m x C solid x  t  = m x C liquid x  t  H =  H vap x # mols  = m x C gas x  t

Heating curves and  H temperature added energy

Heating curves and  H temperature added energy solidliquidgas

Heating curves and  H temperature added energy melting/ freezing pt solidliquidgas

Heating curves and  H temperature added energy melting/ freezing pt boiling/ cond. pt solidliquidgas

Heating curves and  H temperature added energy melting/ freezing pt boiling/ cond. pt solidliquidgas melting/freezing occurring here boiling/condensing occurring here

How is the total enthalpy change (  H) calculated for a substance whose temperature change includes a change in state?

temperature added energy  t of solid absorbing energy

temperature added energy  = m x C solid x  t

temperature added energy  = m x C solid x  t the energy absorbed as a solid melts becomes potential energy, so no  t

temperature added energy  H =  H fus x # mols  = m x C solid x  t

temperature added energy  H =  H fus x # mols  t of liquid absorbing energy  = m x C solid x  t

temperature added energy  H =  H fus x # mols  = m x C liquid x  t  = m x C solid x  t

temperature added energy  H =  H fus x # mols the energy absorbed as a liquid boils becomes potential energy, so no  t  = m x C liquid x  t  = m x C solid x  t

temperature added energy  H =  H fus x # mols  H =  H vap x # mols  = m x C liquid x  t  = m x C solid x  t

temperature added energy  H =  H fus x # mols  H =  H vap x # mols  t of gas absorbing energy  = m x C liquid x  t  = m x C solid x  t

temperature added energy  H =  H fus x # mols  H =  H vap x # mols  = m x C gas x  t  = m x C liquid x  t  = m x C solid x  t

temperature added energy  H =  H fus x # mols  H =  H vap x # mols  = m x C gas x  t  = m x C liquid x  t  = m x C solid x  t

temperature added energy  H =  H fus x # mols  H =  H vap x # mols  = m x C gas x  t  = m x C liquid x  t  = m x C solid x  t The  H of any substance being heated will be the sum of the  H of any  t occurring plus  H of any phase change occurring

temperature added energy  =  H fus x # mols  =  H vap x # mols  = m x C gas x  t  = m x C liquid x  t  = m x C solid x  t The  H of any substance being heated will be the sum of the  H of any  t occurring plus  H of any phase change occurring

temperature added energy  H =  H fus x # mols  H =  H vap x # mols  = m x C gas x  t  = m x C liquid x  t  = m x C solid x  t EXAMPLE : What is  H for 10 g water with a total  t from -20 o C to +50 o C?

temperature added energy -20 o C 0 o C 50 o C EXAMPLE : What is  H for 10 g water with a total  t from -20 o C to +50 o C?

temperature added energy EXAMPLE : What is  H for 10 g water with a total  t from -20 o C to +50 o C? -20 o C 0 o C 50 o C use the following values: C ice = 2.1 J/g o c,  H fus H 2 O = 6.01 kJ/mol, C H 2 O liq = J/g o C

temperature added energy -20 o C 0 o C 50 o C  H  =  H fus x # mols   = m x C liquid x  t   = m x C solid x  t EXAMPLE : What is  H for 10 g water with a total  t from -20 o C to +50 o C? use the following values: C ice = 2.1 J/g o c,  H fus H 2 O = 6.01 kJ/mol, C H 2 O liq = J/g o C

temperature added energy -20 o C 0 o C 50 o C  H  =  H fus x # mols   = 10g x 2.1 J/g o C x 20 o C EXAMPLE : What is  H for 10 g water with a total  t from -20 o C to +50 o C? use the following values: C ice = 2.1 J/g o c,  H fus H 2 O = 6.01 kJ/mol, C H 2 O liq = J/g o C   = m x C liquid x  t

temperature added energy -20 o C 0 o C 50 o C  H  =10 g x 1mol/18g x 6.01kJ/mol   = 10g x 2.1 J/g o C x 20 o C  H 2 =10 g x 1mol/18g x 6.01 kJ/mol EXAMPLE : What is  H for 10 g water with a total  t from -20 o C to +50 o C? use the following values: C ice = 2.1 J/g o c,  H fus H 2 O = 6.01 kJ/mol, C H 2 O liq = J/g o C   = m x C liquid x  t

temperature added energy -20 o C 0 o C 50 o C   = 10g x J/g o C x 50 o C   = 10g x 2.1 J/g o C x 20 o C  H 2 =10 g x 1mol/18g x 6.01 kJ/mol EXAMPLE : What is  H for 10 g water with a total  t from -20 o C to +50 o C? use the following values: C ice = 2.1 J/g o c,  H fus H 2 O = 6.01 kJ/mol, C H 2 O liq = J/g o C

temperature added energy -20 o C 0 o C 50 o C   = 10g x J/g o C x 50 o C   = 10g x 2.1 J/g o C x 20 o C  H 2 =10 g x 1mol/18g x 6.01 kJ/mol EXAMPLE : What is  H for 10 g water with a total  t from -20 o C to +50 o C? use the following values: C ice = 2.1 J/g o c,  H fus H 2 O = 6.01 kJ/mol, C H 2 O liq = J/g o C total  H =   +  H 2 +  

temperature added energy -20 o C 0 o C 50 o C   = 10g x 2.1 J/g o C x 20 o C  H 2 =10 g x 1mol/18g x 6.01 kJ/mol EXAMPLE : What is  H for 10 g water with a total  t from -20 o C to +50 o C? use the following values: C ice = 2.1 J/g o c,  H fus H 2 O = 6.01 kJ/mol, C H 2 O liq = J/g o C total  H =   +  H 2 +   = 420 J J J   = 10g x J/g o C x 50 o C

temperature added energy -20 o C 0 o C 50 o C   = 10g x 2.1 J/g o C x 20 o C  H 2 =10 g x 1mol/18g x 6.01 kJ/mol EXAMPLE : What is  H for 10 g water with a total  t from -20 o C to +50 o C? use the following values: C ice = 2.1 J/g o c,  H fus H 2 O = 6.01 kJ/mol, C H 2 O liq = J/g o C total  H =   +  H 2 +   5853 J = 420 J J J   = 10g x J/g o C x 50 o C

temperature added energy -20 o C 0 o C 50 o C   = 10g x 2.1 J/g o C x 20 o C  H 2 =10 g x 1mol/18g x 6.01 kJ/mol EXAMPLE : What is  H for 10 g water with a total  t from -20 o C to +50 o C? It takes 5853 joules to heat up 10 grams of water from -20 o C to +50 o C.   = 10g x J/g o C x 50 o C

Example II: Ice at 0 o C is placed in a Styrofoam cup containing 0.32 kg of lemonade at 27 o C. The specific heat capacity of lemonade is virtually the same as that of water. After the ice and lemonade reach an equilibrium temperature, some ice still remains. Assume that the cup absorbs a negligible amount of heat. m ice g X 1mol/18g X 6010J/mole = 320g X 4.18J/g.C o X (27 o C -0 o C) m ice = 108.1g