# AP Chemistry Lab Investigation #12 HAND WARMER A solution was formed combining 25.0 g of solid A with 60.0 mL of distilled water, with the water initially at 21.4 degrees celcius. The final...

**AP Chemistry Lab Investigation #12 HAND WARMER **

A solution was formed combining 25.0 g of solid A with 60.0 mL of distilled water, with the water initially at 21.4 degrees celcius. The final temperature of the solution was 25.3 degrees celcius. Calculate the heat released as the solid dissolved, qsoln, assuming no heat loss to the calorimeter.

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### 1 Answer

This is one of those problems that gives you more data than you really need, in order to see if you know which numbers to use in your calculation.

In this question we're fundamentally concerned with a change in temperature. Temperature is really just a way of talking about the kinetic energy of a substance, or how much it's moving around. You can think of the particles as "wiggling", and added energy makes them wiggle even more vigorously.

When solid A is dropped in water, some of its bonds are broken, probably due to water's polarity. The energy stored in these bonds is released, and rather than creating new bonds, it becomes heat, which is just another way of talking about temperature; the surrounding atoms are now wiggling around a bit more energetically than before.

If we were really being thorough, we'd need to evaluate the properties of the solution (solid A + water), but because this isn't provided, we'll just work with the values of water, which should be familiar to you in AP. The **heat capacity of water **is 4.18J/g; this means that 4.18 joules of heat energy will change the temperate of 1 gram of water by 1 degree. Compare this to what we know in this problem;

Initial Temperature: 21.4 C

Final Temperature: 25.3 C

Volume of Water: 60.0mL

Mass of Solid A: 25.0g

To determine the heat that was required to raise the temperature, we need to know the mass of the water that was heated. The mass of Solid A is not relevant to this problem (although you would need it to find the heat capacity of Solid A).

All we really have to do is find out how much mass is in 60.0mL of water. Fortunately water has a pretty convenient conversion: about 1mL per gram. So we should have about 60g of water.

Temperature Change: 25.3 - 21.4 = 3.9 degrees

Mass of Water: 60g

Heat Capacity of Water: 4.18J/g

So 4.18 x 60 = Joules to raise temperature by 1 degree, and

4.18 x 60 x 3.9 = Joules to raise temperature by 3.9 degrees.

4.18 x 60 x 3.9 = **978.12 J**