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.

Answer: 52 J

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

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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