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10 August, 01:39

Suppose that you have 23 g of silver. Also assume that this sample of silver is initially at 18 oC. What is the final temperature of this sample of silver after it absorbs 2.05 kJ of heat? csilver = 0.235 J/goC Enter your answer numerically, in terms of oC.

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  1. 10 August, 02:06
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    396.92°C

    Explanation:

    Hello,

    This question relates to heat transfer between objects and we can simply use the formula to solve this question.

    Data;

    Mass of silver (m) = 23g

    Initial temperature of silver (T1) = 18°C

    Final temperature of silver (T2) = ?

    Heat absorbed = 2.05kJ = 2050J

    Specific heat capacity of silver = 0.235J/g°C

    Heat energy (Q) = mc∇T

    Q = MC∇T

    Q = heat energy

    M = mass of substance

    C = specific heat capacity of substance

    ∇T = change in temperature of substance = (T2 - T1)

    Q = MC∇T

    2050 = 23 * 0.235 * (T2 - 18)

    2050 = 5.41 * (T2 - 18)

    2050 = 5.41T2 - 97.38

    Collect like terms

    2050 + 97.38 = 5.41T2

    2147.38 = 5.41T2

    Divide both sides by 5.41

    2147.38 / 5.41 = 5.41T2 / 5.41

    T2 = 396.92°C

    The final temperature of the sample after absorbing 2.05kJ of heat is 396.92°C

    Have a nice day
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