Suppose that 1.15 g of rubbing alcohol (C3H8O) evaporates from a 67.0 g aluminum

Question

Suppose that 1.15 g of rubbing alcohol (C3H8O) evaporates from a 67.0 g aluminum block. You may want to reference ( pages 502 - 512) Section 11.5 while completing this problem. Part A 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 vaporization of the alcohol comes only from the aluminum block and that the alcohol vaporizes at 25 C. The heat of vaporization of the alcohol at 25 C is 45.4 kJ/mol Express your answer using two significant figures.

Explanation / Answer

Qevap = m*Hvap

Hvap = 45.4 kJ/mol

mol of alcohol = mass/MW = 1.15/60.10 = 0.01913

Qvap = 0.01913*45.4 = 0.868502kJ = 868.5 J

this is lost by aluminium so

Qal = mal*Cpal*(Tf-Tal)

Qal = -868.5

868.5 = 67*0.9*(Tf-25)

Tf = -868.5 /(67*0.9)+25

Tf = 10.60 °C or 11°C (2 sig fig)

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