Calculate the energy, in both joules AND calories, that is required to heat 75 g of water from 2.1 degree C to 86.4 degree C

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calculate the energy, in both joules AND calories, that is required to heat 75 g of water from 2.1 degree C to 86.4 degree C


The freezing point of water is 0C. The boiling point is 100C. Your sample won't go through either of those points, so you don't have to worry about the enthalpy of fusion or the enthalpy of vaporization (if you don't know what those are, you should look them up because your teacher will probably ask a question using those on the test.).

The specific heat of water is 1\,\frac {calories}{g\cdot deg\,C}

Basically, what that means is that it requires 1 calorie to heat 1 gram of water 1 degree Celsius (1 C). If you have two grams, you need two calories, etc. Basically, you multiply this by the number of grams and also by the number of degrees C you're heating the water:

1\,\frac {calories}{g\cdot deg\,C} \times (deg\,C) \times (grams) = calories\,required

You can convert calories to joules rather easily. Click here: Let me google that for you - Calories to Joules (http://www.lmgtfy.com/?q=convert+calories+to+joules&l=1)