Question

The enthalpy change for converting 1.00 mol of ice at -25.0 ∘c to water at 90.0∘c is ________ kj. the specific heats of ice, water, and steam are 2.09 j/g−k, 4.18 j/g−k, and 1.84 j/g−k, respectively. for h2o, δ hfus = 6.01kj/mol, and δhvap = 40.67 kj/mol.

Answer 1

Answer : The enthalpy change for converting 1 mole of ice at $-25.0^oC$ to water at $90^oC$ is, 7.712 KJ

Solution :

Process involved in the calculation of enthalpy change :

$(1):ice(-25^oC)\rightarrow ice(0^oC)\\\\(2):ice(0^oC)\rightarrow water(0^oC)\\\\(3):water(0^oC)\rightarrow water(90^oC)$

Now we have to calculate the enthalpy change.

$\Delta H=[m\times c_{ice}\times (T_2-T_1)]+\Delta H_{fusion}+[m\times c_{water}\times (T_3-T_2)]$

where,

$\Delta H$ = enthalpy change

m = mass of water = $1mole\times 18g/mole=18g$

$c_{ice}$ = specific heat of ice = 2.09 J/gk

$c_{water}$ = specific heat of water = 4.18 J/gk

$\Delta H_{fusion}$ = enthalpy change for fusion = 6.01 KJ/mole = 0.00601 J/mole

conversion : $0^oC=273k$

$T_1$ = initial temperature of ice = $0^oC=273k$

$T_2$ = final temperature of ice = $-25^oC=273+(-25)=248k$

$T_3$ = initial temperature of water = $0^oC=273k$

$T_4$ = final temperature of water = $90^oC=273+90=363k$

Now put all the given values in the above expression, we get

$\Delta H=[18g\times 2.09J/gK\times (273-248)k]+0.00601J+[18g\times 4.18J/gK\times (363-273)k]$

$\Delta H=7712.106J=7.712KJ$     (1 KJ = 1000 J)

Therefore, the enthalpy change for converting 1 mole of ice at $-25.0^oC$ to water at $90^oC$ is, 7.712 KJ