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2 April, 20:40

Suppose during a lab that 65kJ of energy were transferred to 450 g of water at 20 degree Celsius. What would have been the final temperature of the water

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  1. 2 April, 20:50
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    Final temperature of water would have been equal to 54.5 °C.

    Explanation:

    Given:

    Amount of transferred energy = 65 K J

    Mass of water = 450 g

    Initial temperature = 20°C

    To find the final temperature of water:

    Formula for Heat capacity is given by

    Q = m*c*Δt ... (1)

    where:

    Q = Heat capacity of the substance (in J)

    m=mass of the substance being heated in grams (g)

    c = the specific heat of the substance in J / (g.°C)

    Δt = Change in temperature (in °C)

    Δt = (Final temperature - Initial temperature) = T (f) - T (i) ... (2)

    Q = 65 K J,

    As Q is given in K J, to convert kilo-joules into joules, 1 K J = 1000 J

    Therefore 65 K J = 65,000 J

    Q = 65,000 J

    Mass of water = m = 450 g

    Initial temperature of water = T (i) = 20°C

    Specific heat of water is c = 4.186 J / g. °C

    Substituting these in equation (1), we get

    Q = m*c*Δt

    Rearranging the terms for Δt,

    Δt = Q / (m*c)

    Δt = 65,000 / (450*4.186) = 65,000/1883.7

    Δt = 34.5 °C ... (3)

    Substituting (3) in (2) and solving for T (f):

    Δt = T (final) - T (initial) = T (f) - T (i)

    Rearranging the equation for T (f),

    T (f) = Δt + T (i)

    T (f) = 34.5 + 20

    T (f) = 54.5 °C

    Final temperature of water would have been equal to T (f) = 54.5 °C.
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