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18 July, 17:33

The temperature of a sample of iron with a mass of 10.0 g changed from 50.4°C to 25.0°C with the release of 47 calories of heat. What is the specific heat of iron?

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  1. 18 July, 17:57
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    Q = mc (θ₂-θ₁)

    47 calories = 10 g * c * (50.4 - 25)

    47cal = 10*c * 25.4

    47 / (10*25.4) = c

    0.185 = c

    Specific heat of iron = 0.185 cal/g°C
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