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Iron metal has a specific heat of 0.449J/g°C. How much heat is transferred to a 5.00 a piece of iron, initially at 20.0°C, when it is placed in a beaker of boiling water at 1 atm?

Use Q = mcAt formula

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  1. Today, 16:34
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    Q = 44.9 j

    Explanation:

    Mass of iron = 5.0 g

    Change in temperature = 20 °C

    Specific heat of iron = 0.449 j/g.°C

    Heat transferred = ?

    Formula:

    Q = m. c. ΔT

    Q = amount of heat absorbed or released

    m = mass of given substance

    c = specific heat capacity of substance

    ΔT = change in temperature

    Solution:

    Q = m. c. ΔT

    Q = 5 g * 0.449 j/g.°C * 20 °C

    Q = 44.9 j
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