Sandra heated a 12.3-g piece of iron to 100oc. she then placed the iron in a calorimeter with a heat capacity of 10.4 j/oc. the reading on the calorimeter temperature gauge rose from 25oc to 30oc. how much heat did the piece of iron lose? a:         2.08 jb:         52.0 jc:         61.5 jd:         861 j

We first have to find Q calorimeter

Q calorimeter = C (delta T) 

C= heat capacity in J/oC, not small c ( specific heat capacity) 
delta T= change in T (oC) = final T - initial T 

Q calorimeter = 10.4 J/oC (30 oC - 25 oC)
Q calorimeter = 52 J 

Remember, 

Q calorimeter = - Q surrounding
The (-) negative sign stands for heat being released. 

So , 

Q surrounding = - 52 J 

In other words, the iron releases 52 J .

The calorimeter absorbs the entire 52 J and heat rose from 25 - 30 oC. 

52 J