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

is5.02KJ/mol. The specific heat of solid and liquid ethanol are
0.97JK/g-K and 2.31 J/g-K, respectively. How much heat(KJ) is
needed toconvert 25.0 g of solid ethanol at -135 degree C to liquid
ethanolat -50 degree C?

User DaveAlger
by
8.0k points

1 Answer

5 votes

Answer: The heat required is 6.88 kJ.

Step-by-step 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* c_(p,s)* (T_(final)-T_(initial))]+n* \Delta H_(fusion)+[m* c_(p,l)* (T_(final)-T_(initial))]+n* \Delta H_(vap)+[m* c_(p,g)* (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}}=(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* 0.97J/gK* (-114-(-135)K]+0.534mole* 5020J/mole+[25.0g* 2.31J/gK* (-50-(-114))K]


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

Therefore, the heat required is 6.88 kJ

User Zhuo
by
7.8k points
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