Question

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.

Answer 1

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 2

Answer : The enthalpy change is, 6.2315 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)(90^oC)`

Now we have to calculate the enthalpy change.

`\Delta H=[m* c_(p,s)* (T_(final)-T_(initial))]+n* \Delta H_(fusion)+[m* c_(p,l)* (T_(final)-T_(initial))]`

where,

`\Delta H` = enthalpy change = ?

m = mass of ice = 10.0 g

`c_(p,s)` = specific heat of solid water =
`2.09J/g^oC`

`c_(p,l)` = specific heat of liquid water =
`4.18J/g^oC`

n = number of moles of water =
`\frac{\text{Mass of water}}{\text{Molar mass of water}}=(10g)/(18g/mole)=0.55mole`

`\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* 4.18J/gK* (0-(-25))^oC]+0.55mole* 6010J/mole+[10g* 2.09J/gK* (90-0)^oC]`

`\Delta H=6231.5J=6.2315kJ` (1 KJ = 1000 J)

Therefore, the enthalpy change is, 6.2315 KJ