### Time-Temperature Calculation #3:

Liquid Water Warming up

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72.0 grams of liquid water is 0.0 °C. It is going to warm up to 100.0 °C, but at that temperature, the water WILL NOT BOIL. We need to calculate the energy needed to do this.

This summarizes the information needed:

Δt = 100.0 °C

The mass = 72.0 g

C_{p} = 4.184 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 4.184 J for that. Now go the second degree. Another 4.184 J. Go the third degree and use another 4.184 J. So one gram going 100 degress needs 4.184 x 100 = 418.4 J. Now we have 72 grams, so gram #2 also needs 418.4, gram #3 needs 418.4 and so on until 72 grams.

I hope that helped.

With the numbers in place, we have:

q = (72.0 g) (100.0 °C) (4.184 J/g °C)

So we calculate and get 30124.8 J. We won't bother to round off right now since there are two more calculations to go. We will have to do five calculations and then sum them all up.

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 30124.8 J is in kJ.

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