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

whpalmer4 · Answer

The formula relating heat to specific heat, mass and temperature change is 
$$Q = m c \Delta T$$
You know all of the quantities on the right hand side from the problem statement. $\Delta T = \text{final temperature} - \text{starting temperature}$

anonymous · Answer

that is all wrong bruh

whpalmer4 · Answer

@qveengb Oh, really?  Do explain, "bruh"...

tatianagomezb · Answer

I back @whpalmer4  answer. Because when heated iron absorbs energy, that is released by  an unknown source. And using that formula I got 7.7 calories, please correct me if I'm wrong.

anonymous · Answer

um haven't you ever heard of a joke