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We know the specific Heat of water is: 4.18J/(g-k)

So the formula is:

q = Cs * m * change in Temperature

In above equation, q = heat, Cs = Specific heat, m = mass

q = 4.18J/(g-K) * 50 * 15 = 3135

But -3135J because this is an exothermic reaction.

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Q: If a 50 g sample of water is cooled from 30 C to 15 C thereby losing approximately 3140 J of heat what is the change of energy in J?
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