Question

. PLEASE HELP!!! SHOW ALL STEPS THANKS

If 42.0 kJ of heat is added to a 32.0 g sample of liquid methane under 1 atm of pressure at a temperature of -170 ^oC, what are the final state and temperature of the methane once the system equilibrates? Assume no heat is lost to the surroundings. The normal boiling point of methane is -161.5 ^oC. The specific heats of liquid and gaseous methane are 3.48 and 2.22 J/g-K, respectively.

Answer 1

Total heat added, Q 42 kJ mass of methane, m 32 g Initial temperature =-170。C = 103 K boiling point-- 161.5 °C 111.5 K Heat required to raise the temperature upto boiling point 32 x 3.48 x (1115 -103) 946.56 J = 0.94656 kJ moles of methane mass of methane 32g 16g/mol molar mass of methane - 2moles Heat required to change it to vapour form, vap = 2 x 8.20 = 16.4 kJ Let the final temperature be T Heat required to raise the temperature from 111.5 K to T, = 32 x 2.22 x (T-1 11.5) - 0.07104 x (T 111.5 Total heat added = q1 + q2 + q3 42 0.94656 kJ + 16.4 kJ + 0.07104 x (T-111 .5) 42 = 17.34656 + 0.07104 x (T-1115) 24.65344 = 0.07104 x (T-111.5) 347.03 =T- 111.5 T 458.53 K - 185.53°C