Question

If a 38g sample of water releases 621J of heat energy and cools to 4 0C. What was the initial temperature of the water?

Answer 1

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~