### Time-Temperature Calculation #5:

Steam (Water as Gas) Warming up

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72.0 grams of steam is 100.0 °C. It is going to warm up to 120.0 °C. We need to calculate the energy needed to do this.

This summarizes the information needed:

Δt = 20.0 °C

The mass = 72.0 g

C_{p} = 2.02 Joules per gram-degree Celsius

The calculation needed, using words & symbols is:

q = (mass) (Δt) (C_{p})

Why is this equation the way it is?

Think about one gram going one degree. The liquid water needs 2.02 J for that. Now go the second degree. Another 2.02 J. Go the third degree and use another 2.02 J. So one gram going 20 degrees Celsius needs 2.02 x 20 = 44 J. Now we have 72 grams, so gram #2 also needs 44, gram #3 needs 44 and so on until 72 grams.

I hope that helped.

With the numbers in place, we have:

q = (72.0 g) (20.0 °C) (2.02 J/g °C)

So we calculate and get 2908.8 J. We won't bother to round off right now since we still need to sum up all five values.

One warning before going on: three of the calculations will yield J as the unit on the answer and two will give kJ. When you add the five values together, you MUST have them all be the same unit.

In the context of this problem, kJ is the preferred unit. You might want to think about what 2908.8 J is in kJ.

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