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7 January, 02:00

If 980kJ of energy as heat are transferred to 6.2L of water at 291 K what will the final temperature be? The specific heat of water is 4.18 J/g•K. Assume that 1.0 mL of water equals 1.0 g of water.

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  1. 7 January, 02:14
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    Final temperature = T₂ = 328.815 K

    Explanation:

    Given dа ta:

    Given energy = 980 KJ = 980*1000 = 980000 J

    Volume = 6.2 L

    Initial temperature = T₁ = 291 K

    Specific heat of water = 4.18 j / g. K

    Final temperature = T₂ = ?

    Formula:

    Q = m. c. ΔT

    ΔT = T₂ - T₁

    we will first convert the litter into milliliter

    6.2 * 1000 = 6200 mL

    It is given in question that

    1 mL = 1 g

    6200 mL = 6200 g

    Now we will put the values in formula,

    Q = m. c. (T₂ - T₁)

    980000 j = 6200 g. 4.18 j / g. K. (T₂ - 291 K)

    980000 j = 25916 j / k. (T₂ - 291 K)

    980000 j / 25916 j / k = T₂ - 291 K

    37.8145 K + 291 K = T₂

    T₂ = 328.815 K
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