Question

Ethanol melts at -114 degree C. The enthalpy of fusion

Answer 1

Answer: The heat required is 6.88 kJ.

Explanation:

The conversions involved in this process are :

$(1):ethanol(s)(-135^0C)\rightarrow ethanol(s)(-114^0C)\\\\(2):ethanol(s)(-114^0C)\rightarrow ethanol(l)(-114^0C)\\\\(3):ethanol(l)(-114^0C)\rightarrow ethanol(l)(-50^0C)$

Now we have to calculate the enthalpy change.

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

where,

$\Delta H$ = enthalpy change = ?

m = mass of ethanol = 25.0 g

$c_{p,s}$ = specific heat of solid ethanol= 0.97 J/gK

$c_{p,l}$ = specific heat of liquid ethanol = 2.31 J/gK

n = number of moles of ethanol = $\frac{\text{Mass of ethanol}}{\text{Molar mass of ethanol}}=\frac{25.0g}{46g/mole}=0.543mole$

$\Delta H_{fusion}$ = enthalpy change for fusion = 5.02 KJ/mole = 5020 J/mole

$T_{final}-T_{initial}=\Delta T$ = change in temperature

The value of change in temperature always same in Kelvin and degree Celsius.

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

$\Delta H=[25.0 g\times 0.97J/gK\times (-114-(-135)K]+0.534mole\times 5020J/mole+[25.0g\times 2.31J/gK\times (-50-(-114))K]$

$\Delta H=6885.93J=6.88kJ$     (1 KJ = 1000 J)

Therefore, the heat required is 6.88 kJ