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

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

sweetburger · Answer

I think you use Delta Q = mcDeltaT So your heat absorbed is your Q, m is mass, c is specific heat, and Delta T is change in Temperature so just plug in the values