Part A Suppose that 1.00 g of rubbing alcohol (CH,O) evaporates from a 73 0 g al

Question

Part A Suppose that 1.00 g of rubbing alcohol (CH,O) evaporates from a 73 0 g aluminum block If the aluminum block is initially at 25 °C, what is the final temperature of the block after the evaporation of the alcohol? Assume that the heat required for the vaponzation of the alcohol comes You may want to reference Pages 478-487) Section 11.5 while completing this problerm only trom the aluminum block and that the alcohol vaporizes at 25 C The heat of vaporization of the alcohol at 25 Cis 45 4 kJ/mol Express your answer using two significant figures Submit Request Answer

Explanation / Answer

Answer 1

Heat energy required to evaporate the alcohol = heat given by the aluminium block

MaL=mc(t2-t1)

Ma= Mass of alcohol

m= mass of aluminium

c= specific heat capacity of aluminium. (c= 0.900 J/g K )

L= latent heat = 45.4 kJ/mol= 45400 J/mol

So, 1*45400 = 73*0.900*(298-t2)

t2 = -393.02- 273

= 120.02 degree Cesius.

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