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I have a 1.3 kg block of ice that I pull out of the freezer. That...

I have a 1.3 kg block of ice that I pull out of the freezer. That block of ice is initially at a temperature of -20 °C . How much heat is required for this block of ice to turn into 1.3 kg of liquid water at 20 °C.

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Answer #1

Given,

Mass of the block of ice, m=1.3kg

Initial temperature, T_i=-20\degree C

Final temperature, T_f=20\degree C

We know that specific heat capacity of ice, C_i = 2.04kJ kg^{-1}K^{-1}

Latent heat of fusion of ice, l = 3.36*10^5 Jkg-1

Specific heat capacity of water,  C_w = 4.2kJ kg^{-1}K^{-1}

Heat required for the ice to reach 0\degree C ,

  Q_1 = mC_i\Delta T

  Q_1 = 1.3*2.04*10^3*(0-(-20))

  Q_1 = 53040J

Heat required to convert it from ice to water,

  Q_2 = m*l

  Q_2 = 1.3*3.36*10^5

  Q_2 = 436800J

Heat required for water to reach 20\degree C ,

  Q_3=mC_w\Delta T

  Q_3=1.3*4.2*10^3*(20-0)

  Q_3=109200J

Total heat required, Q= Q_1+Q_2+Q_3

  Q= 53040+436800+109200

  Q= 599040J

  Q= 5.99*10^5J

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