# Specific Heat Question

When 13 grams of a sample cools from 56◦C to 34.4◦C it loses 25.8 Joules of heat. What is the specific heat of the sample? Answer in units of J/g ·◦ C.

The specific heat formula is q = mc$\Delta$T, where q is the amount of heat absorbed or released by an object, m is the object's mass, c is the object's specific heat, and $\Delta$T is the change in temperature that the object experiences (in other words, the final temperature minus the initial temperature of the object).