Apiece of unknown metal with mass 68.6 g is heated to an initial temperature of 100 °c and dropped

Apiece of unknown metal with mass 68.6 g is heated to an initial temperature of 100 °c and dropped into 84 g of water (with an initial temperature of 20 °c) in a calorimeter. the final temperature of the system is 52.1°c. the specific heat of water is 4.184 j/g*⁰c. what is the specific heat of the metal?
a) 0.171
b) 0.343
c) 1.717
d) 3.433

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  1. The specific heat of the metal comes out to be [tex]3.423J/g^oC[/tex]

    Explanation: Heat absorbed or heat lost, q is written as:

    [tex]q=mc\Delta T[/tex]

    We are given that an unknown metal is dropped in water. So, in this process:

    Heat lost by the metal = Heat gained by the water

    Mathematically,

    [tex](mc\Delta T)_m=-(mc\Delta T)_w[/tex]     ....(1)

    For metal:

    [tex]T_2=52.1^oC[/tex]

    [tex]T_1=100^oC[/tex]

    [tex]\Delta T=T_2-T-1=(52.1-100)^oC=-47.9^oC[/tex]

    m = 68.6 g

    c = ?

    For Water:

    [tex]T_2=52.1^oC[/tex]

    [tex]T_1=20^oC[/tex]

    [tex]\Delta T=T_2-T-1=(52.1-20)^oC=32.1^oC[/tex]

    m = 84 g

    [tex]c=4.184J/g^oC[/tex]

    Putting this in equation 1, we get

    [tex]68.6g\times c_m\times (-47.9^oC)=84g\times 4.184J/g^oC\times 32.1^oC[/tex]

    [tex]c_m=3.423j/g^oC[/tex]

  2. The heat absorbed by water will be equal to heat given by the metal

    The heat absorbed by water = Q = msΔt

    Where

    m = mass of water

    s = specific heat of water = 4.18 J g⁻¹°C⁻¹

    Δt = final temperature - initial temperature = (52.1-20)°C = 32.1°C

    So heat absorbed by water = 84 X 4.184 X 32.1 = 11281.74 J

    Heat given by metal

        = mass of metal X specific heat of metal X change in temperature

    heat given by metal = 68.6 X specific heat of metal X (100-52.1)

    11281.74 J = 68.6 X specific heat of metal X (100-52.1)

    specific heat of metal = 3.43 J g⁻¹°C⁻¹

  3. 52.1-4.184=47.916 You then either add or subtract this from 68.6 the first temperature of the metal. This will give you your total.

  4. Option (d)

    Explanation:

    Let c be the specific heat of the metal.

    By use of principle of caloriemetry,

    Heat lost by the hot body = Heat gained by the cold body

    Heat lost by metal = Heat gained by water

    mass of metal x specific heat of metal x fall in temperature of metal = mass of water x specific heat of water x rise in temperature of water

    68.6 x c x (100 - 52.1) = 84 x 4.184 x (52.1 - 20)

    3285.94 x c = 11281.7376

    c = 3.433 J/g C

  5. The piece of unknown metal is in thermal equilibrium with water such that Q of metal is equal to Q of the water. We write this equality as follows:

    -Qm = Qw
    Mass of metal (Cm)(ΔT) = Mass of water (Cw) (ΔT)

    where C is the specific heat capacities of the materials.

    We calculate as follows:

    -(Mass of metal (Cm)(ΔT)) = Mass of water (Cw) (ΔT)
    -68.6 (Cm)(52.1 - 100) = 42 (4.184) (52.1 - 20)
    Cm = 1.717 > OPTION C

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