The answer would be D because a chemical property is Burning is a chemical property and a chemical property is any of material's properties that becomes evident during or after a chemical reaction that is any quality that can be established only by changing a substance's chemical identity.
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Answer : The normal boiling point of ethanol will be,
or 
Explanation :
The Clausius- Clapeyron equation is :

where,
= vapor pressure of ethanol at
= 98.5 mmHg
= vapor pressure of ethanol at normal boiling point = 1 atm = 760 mmHg
= temperature of ethanol = 
= normal boiling point of ethanol = ?
= heat of vaporization = 39.3 kJ/mole = 39300 J/mole
R = universal constant = 8.314 J/K.mole
Now put all the given values in the above formula, we get:


Hence, the normal boiling point of ethanol will be,
or 
U won’t get hyperthermia or a heat stroke when ur hot u use up more energy but when ur cold u use up less