Question

2. (10 pts) If 56.0 k) of heat is added to a 41.0-g sample of liquid temperature of -170 °C, what is the final state final temperature once the system equilibrates? 108 sample of liquid methane under 1 atm of pressure at a methane once the system equilibrates? What is the Assume no heat is lost to the surroundings. The normal boiling point of methane is -161.5.C. AHvap = 8.20 * The specific heats of liquid and gaseous methane are cats of liquid and gaseous methane are 3.48 and 2.22 J/g.K, respectively.

Answer 1

A.

Heat required to increase the temperature from -170^oc to -161.5⁰c

= (mass × specific heat × $\Delta$ T )

= 41 × 3.48 × [ -161.5 -(-170)]

= 1213 J.

= 1.213 KJ.

B.

Moles of methane = (mass/molar mass) = (41/16) = 2.56

Heat required to convert liquid methane to gaseous methane

= Moles of methane× heat of fusion

= 2.56 × 8.20

= 21.01 KJ.

C.

Now heat remaining

= 56.0 - (1.213+21.01) = 33.77 KJ = 33.77×1000 = 33770 J

Now

H = mass × specific heat of gaseous methane × [T_f - (-161.5)]

Or, ( 33770 ) = 41× 2.22 × (T_f +161.5)

Or, (T_f + 161.5 ) = 371

Or, T_f = 371 - 161.5 = 209.5 ⁰c .

Hence, final state is gaseous methane and temperature is 209.5 ⁰c.