A 100.0-g sample of water at 27.0oC is poured into a 71.0-g sample of water at 89.0oC. What will be the final temperature of the water? (Spe

Question

A 100.0-g sample of water at 27.0oC is poured into a 71.0-g sample of water at 89.0oC. What will be the final temperature of the water? (Specific heat capacity of water = 4.184 J/goC.)

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Bình An 5 years 2021-07-18T09:53:44+00:00 1 Answers 52 views 0

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  1. mocmien2
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    2021-07-18T09:55:31+00:00
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    Answer: The final temperature will be 52.74^oC

    Explanation:

    Calculating the heat released or absorbed for the process:

    q=m\times C\times (T_2-T_1)

    In a system, the total amount of heat released is equal to the total amount of heat absorbed.

    q_1=-q_2

    OR

    m_1\times C\times (T_f-T_1)=-m_2\times C\times (T_f-T_2) ……(1)

    where,

    C = heat capacity of water = 4.184J/g^oC

    m_1 = mass of water of sample 1 = 100.0 g

    m_2 = mass of water of sample 2 = 71.0 g

    T_f = final temperature of the system = ?

    T_1 = initial temperature of water of sample 1 = 27^oC

    T_2 = initial temperature of the water of sample 2 = 89.0^oC

    Putting values in equation 1, we get:

    100.0\times 4.184\times (T_f-27)=-71.0\times 4.184\times (T_f-89)\\\\171T_f=9019\\\\T_f=\frac{9019}{171}=52.74^oC

    Hence, the final temperature will be 52.74^oC

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