Calorimetry a. How much heat needs to be taken from 0.5 kg of ethyl alcohol as a

Question

Calorimetry

a. How much heat needs to be taken from 0.5 kg of ethyl alcohol as a gas at its boiling point of 78.00 C to a solid at its freezing point -1140 C if the latent heat of vaporization is 850,000 J/kg and its latent heat of fusion is 104,000 J/kg, and the specific heat is 2400 J/(kg C).

b. If an piece of Iron (c = 450 J/kg C) at 65.00 C is dropped into .250 kg of ethyl alcohol at 25.00 C and the final temperature of the two comes to 35.00 C what was the mass of the piece of iron?

c. If .35 kg of solid ethyl alcohol at -114.00 C is put in a large well insulated jug holding 2.5 kg of liquid ethyl alcohol initially at 35.00 C what will the final temperature be after all the solid form has melted and come to equilibrium?

Explanation / Answer

(a) Total amount of heat = nCdT = 0.5*10*2400*(-1140-78)/46.06844

= -317267.0922

and latent heat = 850000+104000 = 954000

So, amount of heat taken = -317267.0922+954000 = 636732.9 J/kg

(b) since the amount of heat is equal,

0.250*10^3*2400*(65-25)/46.0684 - x*10^3*450*(65-25)/ 55.845 = 10^3*(0.250)*(2400)*(35-25)/46.0684 - 10^3*(x)*(450)*(35-25)/ 55.845

=> 520964.478 - x(322320.709) = 130241.119 - x(80580.177)

or x = 1.616 kg

(c) 0.35*10^3*2400*(-114.00-T)/46.0684 = 2.5*10^3*2400*(35-T)/46.0684

=> 0.35(-114-T) = 2.5(35-T)

=> -39.90-0.35T = 87.5-2.5T

or 2.15T = 126.5 or T = 58.837 C

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