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19 November, 01:55

The specific heat of iron is 0.107 cal/g x oC. How much heat is transferred when a 24.7 kg iron ingot is cooled from 880 oC to 13 oC?

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  1. 19 November, 05:51
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    heat = 229139.43 cal

    Explanation:

    given data

    specific heat of iron is 0.107 cal/g-°C C = 0.4494 J/g°C;

    Mass of iron = 24.7 kg = 24700 g

    temperature T1 = 880°C

    temperature T2 = 13°C

    solution

    we know that Heat lost that is express as

    heat loss = mcΔT ... 1

    here m is the mass and c is specific heat capacity of iron and ΔT is the change in temperature

    so here change in the temperature is = 880°C - 13°C

    change in the temperature = 867°C

    so put here value and we get

    heat = 24700 * 0.107 cal/g-°C * 867°C

    heat = 229139.43 cal
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