Question

How much heat (in kJ) is required to warm 12.0 g of ice, initially at -12.0 °C, to steam at 109.0 °C? The heat capacity of ice is 2.09 J/g⋅°C and that of steam is 2.01 J/g⋅°C.

Please help! Please show work if posible. Thank you!

Answer 1

12 grams of ice from -12 to 109 degree celcius

-12 ==> 0 ==> 100 ==> 109 degree celcius

Important Specific heat capacity of ice = 2.01 J/g⋅°C

  Specific heat capacity of H2O = 4.186J/g⋅°C

Latent heat of vaporization of water = 2260 J/g

Latent heat of fusion of water = 334 J/g

Specific heat capacity of steam = 1.84 J/g⋅°C

Q1 = Heat needed to change temperature from -12 to 0°C = 12 x 2.01 x (0 - (-12°C)) J = 289.44 J

Q2 = Heat needed to change the phase from ice at 0°C to water at 0°C = 12 x 334 J = 4008 J

Q3 = Heat needed to change temperature of water from 0 to 100°C = 12 x 4.186 x (100-0°C) J = 5023.2 J

Q4 = Heat needed to change the phase from water at 100°C to steam at 100°C = 12 x 2260 J = 27120 J

Q5 = Heat needed to change temperature of steam from 100 to 109°C = 12 x 1.84 x (109-100°C ) J = 198.72 J

Total heat absorbed = Q1 + Q2 + Q3 + Q4 + Q5 = 36639.36 J = 36.639 kJ

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