Question

Calculate the amount of heat needed to boil 183.g of ethanol ( CH3CH2OH ), beginning from a temperature of 33.9°C . Round your answer to 3 significant digits. Also, be sure your answer contains a unit symbol.

Answer #1

Sol :-

Q = m x c x dT

mass of ethanol = 183 g

heat capacity of ethanol = 2.238 J g^{-1}
K^{-1}

dT = Change in temperature = T_{f} - T_{i} =
78.3 ^{0}C (B.P of ethanol) - 33.9 ^{0}C = 44.4
^{0}C or **44.4 K**

= 183 g x 2.238 J g^{-1} K^{-1} x 44.4 K

= 18184.1976 J

= 18.2 KJ

= **18.2 KJ**

Moles of ethanol = 183 g/ 46 g/ moles

= **3.98 mol**

Now heat of vaporization = 38.56 KJ/ mol x number of moles

= 38.56 KJ/ mol x 3.98 mol

= **153.5 KJ**

Total heat = 18.2 KJ + 153.5 KJ

**= 171.7 KJ**

**Hence, Total heat needed = 171.7 KJ**

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Please Help

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