If a 38g sample of water releases 621j of heat energy and cools to 4 0c. what was the initial temperature of the water?

Let x be the initial temperature of the water.

According to calorimetry formula, we have:
Q=m*c*ΔT
⇒621J= (38g)*(4.18J/(C*g))*(x-4)C
⇒ x= 8 degrees C 
 
Hope this would help~
       

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