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19 August, 20:57

Determine the specific heat of a 213.0 g piece of metal in J/g•°C if the temperature of the metal drops from 70.2 °C to 25.6 °C when placed in a calorimeter with 250.0 mL of water at an initial temperature of 22.5 °C (specific heat of water is 4.184 J/g•K).

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  1. 20 August, 00:51
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    0.3415 J/g °C

    Explanation:

    M1 = 213.0g

    c1 = ?

    T1 = 70.2°C

    T3 = 25.6 °C

    T2 = 22.5°C

    C2 = 4.186 J/g°C

    Vol. of water = 250mL = 250cm³

    p (density of water) = 1g/L

    p = mass / volume

    mass (m) = density (p) * volume (v)

    m = 1 * 250 = 250g

    Heat loss by metal = heat gain by water

    Q1 = Q2

    Mc∇T = Mc∇T

    Mc (T1 - T3) = Mc * (T3 - T2)

    213 * c * (70.2 - 25.6) = 250 * 4.186 * (25.6 - 22.5)

    (213 * 44.6) c = 1046.5 * 3.1

    9499c = 3244.15

    c = 0.3415 J/g.°C

    The specific heat capacity of the metal is 0.3415 J/g °C
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