Question

find the specific heat of a material that lost about 2,522.616j of energy in the calorimeter if it has a mass of 72.1g and experienced a change in temperature from 100°c to 24°c?

Answer 1

The equation for specific heat is: Q = mc(T2 - T1), and the given shows Q = -2522.616 J (negative because heat is lost), m = 72.1 g, T2 = 24 degrees C, T1 = 100 degrees C. We can solve for specific heat c:
-2522.616 = 72.1(c)(24 - 100) = 0.4604 J/g-degrees C or 0.4604 J/g-K.