How much heat is evolved in converting 1.00 mol of steam at 145.0 ∘C to ice at -55.0 ∘C? The heat capacity of steam is 2.01 J/(g⋅∘C) and of ice is 2.09 J/(g⋅∘C).

asked by @student63603 • about 1 year ago • Chemistry • 5 pts
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Hey there. So this question is similar to one from earlier. You have to go through the different phase changes experienced by water as it moves through different temperatures.

From 145.0 degrees Celsius to 100 degrees Celsius it exists as steam and q = m x c x delta T.

q1 = m x c x delta T = (18.0 g) (2.01 J / g x C) (100 - 145 C) = -1628.1 J

At 100 degrees Celsius steam undergoes a phase change to condense into liquid water.

q2 = m x delta Hcondensation = (18.0 g) (-2260 J/g) = - 40,680 J

From 100.0 degrees Celsius to 0 degrees Celsius it exists as a liquid and q = m x c x delta T.

q3 = m x c x delta T = (18.0 g) (4.184 J/ g x C) (0 -100 C) = -7531.2 J

At 0 degrees Celsius liquid water undergoes a phase change to freeze into solid ice.

q4 = m x delta H = (18.0 g) (-334 J /g ) = -6012 J

From 0.0 degrees Celsius to -55.0 degrees Celsius it exists as a solid and q = m x c x delta T.

q5 = m x c x delta T = (18.0 g) (2.01 J/ g x C) (-55.0 - 0 C) = -1989.9 J

Now if you add all 5 values for q then you will get the overall energy released in the process. The answer should be a negative value because releasing heat is exothermic. However, for some online homework they may not care about the sign and be fine with the answer being entered as a positive value so be careful. Also please check the heat capacity of steam because most literature has it closer to 1.84 J/ g x C instead of 2.01.

Hope that all helps.

answered by @jules • about 1 year ago