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11 May, 07:21

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 J B: 52.0 J C: 61.5 J D: 861 J

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  1. 11 May, 08:16
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    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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