Question

Suppose a 14-gram sample of iron is heated from 20.0°C to 25.0°C. The specific heat of iron is 0.11 cal/g°C. How much heat energy was absorbed by the iron?

38.5 cal

7.7 cal

636 cal

69.3 cal

Answer 1

can't really stay long but you just gotta use

q = mC * deltaT
where m is the mass, C is the specific heat, and deltaT is the temperature change (so final temperature - initial temperature)

Answer 2

Okay - thank ya again!