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11 April, 01:15

The specific heat of aluminum is 0.0215 cal/g°c. if a 4.55 g sample of aluminum absorbs 2.55 cal of energy, by how much will the temperature of the sample change?

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  1. 11 April, 01:41
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    The formula we will use on this one is:

    E = m C ΔT

    where,

    E = energy

    m = mass of sample

    C = specific heat

    ΔT = change in temperature

    Calculating for ΔT:

    ΔT = E / m C

    ΔT = 2.55 cal / (4.55 g * 0.0215 cal/g°C)

    ΔT = 0.098 °C
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