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Ymorist [56]
3 years ago
13

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
Chemistry
1 answer:
Thepotemich [5.8K]3 years ago
8 0
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. 

Answer: 52 J


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