You place an ice cube of mass 7.50×10−3kg and temperature 0.00∘C on top of a copper cube of mass 0.540 kg. All of the ice melts, and the fin

Question

You place an ice cube of mass 7.50×10−3kg and temperature 0.00∘C on top of a copper cube of mass 0.540 kg. All of the ice melts, and the final equilibrium temperature of the two substances is 0.00∘C. What was the initial temperature of the copper cube? Assume no heat is exchanged with the surroundings.

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Calantha 2 weeks 2021-07-21T21:27:42+00:00 1 Answers 0 views 0

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    2021-07-21T21:28:43+00:00

    Answer:

    The value is T_c  =  12 .1 ^oC

    Explanation:

    From the question we are told that

    The mass of the ice cube is m_i  =  7.50 *10^{-3} \  kg

    The temperature of the ice cube is T_i = 0^o C

    The mass of the copper cube is m_c  =  0.540 \  kg

    The final temperature of both substance is T_f  =  0^oC

    Generally form the law of thermal energy conservation,

    The heat lost by the copper cube = heat gained by the ice cube

    Generally the heat lost by the copper cube is mathematically represented as

    Q =  m_c  *  c_c *  [T_c  -  T_f ]

    The specific heat of copper is c_c  = 385J/kg \cdot  ^oC

    Generally the heat gained by the ice cube is mathematically represented as

    Q_1 =  m_i * L

    Here L is the latent heat of fusion of the ice with value L  =  3.34 * 10^{5} J/kg

    So

    Q_1 =  7.50 *10^{-3} * 3.34 * 10^{5}

    => Q_1 =  2505 \ J

    So

    2505  =  0.540  *  385 *  [T_c  - 0 ]

    =>    T_c  =  12 .1 ^oC

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