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Converting J, calories, Calories, and KJ

1.) How much heat is absorbed when 132.2g of water is heated from 12.0 degrees Celsius to 85.0 degrees Celsius. Give your heat change in Joules and kilojoules.
2.) Specific Heat for water is 4.18 J/g Celsius degrees. How much heat is absorbed when 16.8g of water is heated from 3.0 degrees Celsius to 99.0 degrees Celsius. Give your heat change in Joules and calorie.

To convert from Joules to Calories use "dimensional analysis":

$\frac {Joules}{1} \times \frac {X\,Calories}{Joule} = Calories$

Note how Joules cancels out. The conversions are similar for the other problems.

As for the actual conversion factor, Let me google that for you

Quote:

1.) How much heat is absorbed when 132.2g of water is heated from 12.0 degrees Celsius to 85.0 degrees Celsius. Give your heat change in Joules and kilojoules.

When water is heated from 12 to 85 degrees C, the temperature change is (85-12)=73 C. The specific heat of water is 1 cal/(gram C). There are no phase changes (solid to liquid or liquid to gas) in this temperature region, so you don't have to add the latent heat of vaporization or the latent heat of fusion.

So

$\frac {(x\,gram)\, (\Delta T \, Celsius)}{1} \frac {1\,calories}{gram\,Celsius} =1 x \Delta T\,calories$

Note how the Celsius (degrees Celsius) and the grams cancel, leaving calories.

Quote:

2.) Specific Heat for water is 4.18 J/g Celsius degrees. How much heat is absorbed when 16.8g of water is heated from 3.0 degrees Celsius to 99.0 degrees Celsius. Give your heat change in Joules and calorie.

Here the specifie heat is given in J/(g C). The problem is the same. ∆T = (99-3)=96 Celsius. The weight of water is given as 16.8 grams.

$\frac {(x\,gram)\, (\Delta T \, Celsius)}{1} \frac {4.18\,Joules}{gram\,Celsius} =4.18 x \Delta T\,Joules$

How to convert kJ to calories