In a container of negligible mass, 020 kg of ice at an initial temperature of – 40.0 oC is mixed with a mass m of water that has an initial

Question

In a container of negligible mass, 020 kg of ice at an initial temperature of – 40.0 oC is mixed with a mass m of water that has an initial temperature of 80.0 oC. No heat is lost to the surroundings. If the final temperature of the system is 20.0 oC, what is the mass m of the water that was initially at 80.0 oC

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RuslanHeatt 5 years 2021-08-04T14:23:28+00:00 1 Answers 21 views 0

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  1. thuhuong
    0
    2021-08-04T14:25:00+00:00
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    Answer:

    The mass is  m_w = 0.599 \ kg

    Explanation:

    From the question we are told that    

         The mass of ice is  m_c = 0.20 \ kg

         The  initial temperature of the ice is  T_i = -40.0 ^oC

         The  initial temperature of the water is  T_{iw} = 80^o C

         The  final temperature of the system is T_f = 20^oC

    Generally according to the law of energy conservation,

       The  total heat loss is  =  total heat gained

     Now the total heat gain is mathematically represented as

          H = H_1 + H_2 + H_3

    Here  H_1 is the energy required to move the ice from -40^oC \to 0^oC

    And it mathematically evaluated as

         H_1 = m_c * c_c * \Delta T

    Here the specific heat of ice is  c_c = 2100 \ J \cdot kg^{-1} \cdot ^oC^{-1}

    So  

        H_1 = 0.20 * 2100 * (0-(-40))  

         H_1 = 16800\ J

    H_2 is the energy to melt the ice

    And it mathematically evaluated as

           H_2 = m * H_L

    The  latent heat of fusion of ice is  H_L = 334 J/g = 334 *10^{3} J /kg

    So  

        H_2 = 0.20 * 334 *10^{3}

        H_2 = 66800 \ J

    H_3 is the energy to raise the melted ice to 20^oC

    And it mathematically evaluated as

        H_3 = m_c * c_w * \Delta T

    Here the specific heat of water  is  c_w= 4190\ J \cdot kg^{-1} \cdot ^oC^{-1}

        H_3 = 0.20 * 4190* (20-0))  

         H_3 = 16744 \ J  

    So

      H = 16800 + 66800 + 16744

       H = 100344\ J

    The  heat loss is mathematically evaluated as

         H_d = m * c_h ( 80 - 20 )

         H_d = m_w * 4190 * ( 80 - 20 )

         H_d = 167600 m_w

    So

          167600 m_w = 100344

    => m_w = 0.599 \ kg

         

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