How much heat energy must be removed from 150 g of oxygen at 22$\degree$C to liquefy it at -183$\degree$C? (The specific heat of oxygen gas is 0.218 cal/g·$\degree$C, and its heat of vaporization is 50.9 cal/g.)
A) 14,300 cal
B) 9560 cal
C) 4320 cal
D) 2160 cal

Question

Quizplus · Accepted Answer

Explanation:The heat energy that must be removed to liquefy the oxygen is the sum of the heat energy required to cool the gas to its liquefaction temperature and the heat of vaporization.Heat energy to cool the gas:Q1 = m * c * (T2 - T1)= 150 g * 0.218 cal/g·°C * (–183°C – 21°C)= 6,540 calHeat of vaporization:Q2 = m * L= 150 g * 50.9 cal/g= 7,760 calTotal heat energy:Q = Q1 + Q2= 6,540 cal + 7,760 cal= 14,300 cal

How Much Heat Energy Must Be Removed from 150 G °\degree°

How Much Heat Energy Must Be Removed from 150 G $\degree$