Chemistry: The enthalpy change for converting 10.0 g of ice at -25.0°c to water at 80.0°c is kj. the specific heats of ice, water, and steam are 2.09 j/gk, 4.18 j/gk, and 1.84 j/gk, respectively. for h2o, δhfus = 6.01 kj/mol, and δhvap = 40.67 kj/mol.

The enthalpy change for converting 10.0 g of ice

haileymaree

14.03.2022

Answer : The enthalpy change is, 7.205 KJ

Solution :

The conversions involved in this process are :

$(1):H_2O(s)(-25^oC)\rightarrow H_2O(s)(0^oC)\\\\(2):H_2O(s)(0^oC)\rightarrow H_2O(l)(0^oC)\\\\(3):H_2O(l)(0^oC)\rightarrow H_2O(l)(80^oC)$

Now we have to calculate the enthalpy change.

$\Delta H=[m\times c_{p,s}\times (T_{final}-T_{initial})]+n\times \Delta H_{fusion}+[m\times c_{p,l}\times (T_{final}-T_{initial})]$

where,

$\Delta H$ = enthalpy change = ?

m = mass of water = 10 g

$c_{p,s}$ = specific heat of solid water = 2.09 J/gk

$c_{p,l}$ = specific heat of liquid water = 4.18 J/gk

n = number of moles of water = $\frac{\text{Mass of water}}{\text{Molar mass of water}}=\frac{10g}{18g/mole}=\frac{10}{18}mole$

$\Delta H_{fusion}$ = enthalpy change for fusion = 6.01 KJ/mole = 6010 J/mole

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

$\Delta H=[10g\times 2.09J/gK\times (273-248)k]+\frac{10}{18}mole\times 6010J/mole+[10g\times 4.18J/gK\times (353-273)k]$

$\Delta H=7205.39J=7.205KJ$ (1 KJ = 1000 J)

Therefore, the enthalpy change is, 7.205 KJ

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