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Ryan places 0.150 kg of boiling water in a thermos bottle. How many kgs of ice at –12.0 °C must Ryan add to the thermos so that the equilibrium temperature of the water is 75?a. 0.0436 kgb. 0.0713 kgc. 0.0233 kgd. 0.0265 kge. 0.625 kg

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okpalawalter8

Answer:

The  value  is   [tex] m_i =  0.0234 \  kg   [/tex]

Explanation:

Generally from the calorimetry principle we have that

      [tex]Heat \ loss\ by\ ice  =  heat\ gained\ by\  water\[/tex]

So here heat gained water is mathematically represented as i.e

      [tex]Q_w  = m_w  *  c_w *  (T_w - T )[/tex]

substituting  0.150 kg for [tex]m_w [/tex] , 4200 J/kg.°C for  [tex]c_w [/tex] , 100°C for   [tex]T_w [/tex] and  75°C for  T

We have  

       [tex]Q_w  = 0.150  *  4200 *  (100 - 75 )[/tex]

         [tex]Q_w  =15750 \  J [/tex]

The Heat loss by the ice is mathematically represented as

      [tex]Q_i  = Q_1 + Q_2 +  Q_3[/tex]

Here     [tex]Q_1[/tex] is the energy to move the ice to its melting point which is evaluated as  

        [tex]Q_1  =  m_i *  c_i * ( T_o -T_i)[/tex]

Here  [tex]m_i [/tex] is the mass of  ice

       [tex]c_i [/tex] is the specific heat of ice with value  [tex]2.05 * 10^3   J/kg.°C[/tex]

          [tex]T_o [/tex] temperature of ice at melting point with value 0°C

           [tex]T_i [/tex] is the temperature of ice with value  -12°C

[tex]Q_2[/tex] is the energy to move the ice from its  its melting point to liquid which is evaluated as  

     [tex]Q_2  = m_i  *  L[/tex]

        Here  L  is the Latent heat of melting of ice with value    [tex]334 * 10^3   J/kg [/tex]

[tex]Q_3[/tex] is the energy to move the ice from  liquid  to  the equilibrium temperature  which is evaluated as        

       [tex]Q_1  =  m_i *  c_w * ( T -T_o)[/tex]

So  

     [tex]Q_i  = m_i [ c_i * ( T_o -T_i) + L  + c_w * ( T -T_o) ] [/tex]

=>   [tex]Q_i  = m_i [ 2.05 * 10^3 * ( 0 -(-12)) + 334 * 10^3  +  4200 * ( 75 - 0) ] [/tex]

From

 [tex]Heat \ loss\ by\ ice  =  heat\ gained\ by\  water\[/tex]

We have that

  [tex] m_i *  673600  =15750  [/tex]

=>     [tex] m_i =  \frac{15750}{673600}  [/tex]

=>     [tex] m_i =  0.0234 \  kg   [/tex]

Histrionicus

The amount of ice at –12.0 °C must Ryan add to the thermos so that the equilibrium temperature of the water is 75 - 0.0233 kg

The calorimetry principle we have that

  heat is given by water   =   heat is received by ice

=>  

Given:

mass of boiling water [tex]m_w = 0.150 kg[/tex]

temperature of ice [tex]T_i = - 12^0C[/tex]

Known:

  • specific heat of water  [tex]c_w = 4.2 \times 10^3 J/kg-^0C[/tex] the Initial temperature of boiling water   [tex]T_w = 100^0C[/tex]
  • mass of ice  =   [tex]m_i[/tex]
  • specific heat of ice   [tex]c_i = 2.05 \times 10^3 J/kg-^0C[/tex] Latent heat of melting of ice   [tex]L = 334 \times 10^3 J/kg[/tex]   Melting point of ice  [tex]T = 0^0C[/tex] Equilibrium temperature   T   =   75 C

Solution:

placing value in the formula:

=>  

=>

=   0.0233 kg

Thus, the amount of ice at –12.0 °C must Ryan add to the thermos so that the equilibrium temperature of the water is 75 - 0.0233 kg

Learn more:

https://brainly.com/question/14398692

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