Question

How much heat energy, in kilojoules, is required to convert 72.0 gg of ice at −−18.0 ∘C∘C to water at 25.0 ∘C∘C ? Express your answer to three significant figures and include the appropriate units. View Available Hint(s)

Answer 1

Answer: The enthalpy change is 34.3 kJ

Explanation:

The conversions involved in this process are :

$(1):H_2O(s)(-18^0C)\rightarrow H_2O(s)(0^0C)\\\\(2):H_2O(s)(0^0C)\rightarrow H_2O(l)(0^0C)\\\\(3):H_2O(l)(0^0C)\rightarrow H_2O(l)(25^0C)$

Now we have to calculate the enthalpy change.

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

where,

$\Delta H$ = enthalpy change = ?

m = mass of water = 72.0  g

$c_{s}$ = specific heat of ice = $2.09J/g^0C$

$c_{l}$ = specific heat of liquid water = $4.184J/g^0C$

n = number of moles of water = $\frac{\text{Mass of water}}{\text{Molar mass of water}}=\frac{72.0g}{18g/mole}=4.00moles$

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

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

$\Delta H=[72.0g\times 2.09J/g^0C\times (0-(-18)^0C]+4.00mole\times 6010J/mole+[72.0g\times 4.184J/g^)C\times (25-0)^0C]$$\Delta H=34279.8J=34.3kJ$        (1 KJ = 1000 J)

Therefore, the enthalpy change is 34.3 kJ