The enthalpy change for converting 10.0 g 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, = 6.01 kJ/mol, and = 40.67 kJ/mol.

The total heat considering both the sensible and latent heat is expressed by ΔH = (m Cp ΔT)solid + (m ΔHv) + (m Cp ΔT)liquid.
Subsituting the data to the equation,
ΔH = (10 g *2.09 J/g K *25 K) + (10 g *80 cal/g *4.1858 J/cal) + (10 g* 4.18 J/gK * 90 K)
ΔH = 7633.14 Joules

Answer Link

Alleei Alleei · Answer 2 · 2019-09-24T09:56:12+02:00

Answer : The enthalpy change is, 6.2315 KJ

Solution :

The conversions involved in this process are :

[tex](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)(90^oC)[/tex]

Now we have to calculate the enthalpy change.

[tex]\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})][/tex]

where,

[tex]\Delta H[/tex] = enthalpy change = ?

m = mass of ice = 10.0 g

[tex]c_{p,s}[/tex] = specific heat of solid water = [tex]2.09J/g^oC[/tex]

[tex]c_{p,l}[/tex] = specific heat of liquid water = [tex]4.18J/g^oC[/tex]

n = number of moles of water = [tex]\frac{\text{Mass of water}}{\text{Molar mass of water}}=\frac{10g}{18g/mole}=0.55mole[/tex]

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

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

[tex]\Delta H=[10g\times 4.18J/gK\times (0-(-25))^oC]+0.55mole\times 6010J/mole+[10g\times 2.09J/gK\times (90-0)^oC][/tex]

[tex]\Delta H=6231.5J=6.2315kJ[/tex] (1 KJ = 1000 J)

Therefore, the enthalpy change is, 6.2315 KJ