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29 May, 12:29

A 125G sample of water was heated to 100.0°C and then I borrow platinum 20.0°C is dropped into the beaker the temperature of the platinum in the beaker quickly rose 235.0°C the specific heat of platinum is 0.13 j/g°C. The specific heat of water is 4.184 J/g°C. What is the mass of platinum

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  1. 29 May, 14:03
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    mass of platinum = 2526.12 g

    Explanation:

    Given dа ta:

    Mass of water = 125 g

    Initial temperature of water = 100.0°C

    Initial temperature of Pt = 20.0°C

    Final temperature = 235°C

    Specific heat of Pt = 0.13 j/g°C

    Specific heat of water = 4.184 j/g°C

    Mass of platinum = ?

    Solution:

    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

    ΔT = T2 - T1

    Q (w) = Q (Pt)

    m. c. (T2 - T1) = m. c. (T2 - T1)

    125 g * 4.184 j/g°C * (235°C - 100.0°C) = m * 0.13 j/g°C * (235°C - 20°C)

    125 g * 4.184 j/g°C * 135°C = m * 0.13 j/g°C * 215°C

    70605 j = m*27.95 j/g

    m = 70605 j / 27.95 j/g

    m = 2526.12 g
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