The heat required to convert 1 gram of ice at -30 degree C to steam at 120 degre

Question

The heat required to convert 1 gram of ice at -30 degree C to steam at 120 degree C. The heat required to change the temperature of the ice from -30 degree C rightarrow 0 degree C, no phase change, is Q = mc Delta t. Q = (mc Delta t) ice = (10-3kg)(2090 J / kg degree C)(30 degree C) = 62.7 J Total heat added = 62.7 Joules When the ice reaches 0 degree C it begins to melt. The ice/water mixture remains at 0 degree C even though heat is being added because of the phase change. This continues until all of the ice melts. The heat required for the mixture to complete the phase change is Q = mLf. Q = mLf = (10-3 kg)(3.33 Times 105 J/kg) = 333 J Total heat added = 395.7 Joules The heat required to change the temperature of the water from 0 degree C rightarrow 100 degree C, no phase change, is Q = mc Delta t. Q = (mc Delta t)water = (10-3kg)(4.19 Times 103 J/kg degree C)(100 degree C) = 419 J Total heat added = 813.7 Joules When the water reaches 0 degree C it begins to boil. The water/steam mixture remains at 100 degree C even though heat is being added because of the phase change. This continues until all of the water is converted to steam. The heat required for the mixture to complete the phase change is Q = mLv. Q = mLv = (10-3 kg)(2.26 Times 106 J/kg) = 2260 J Total heat added = 3073.7 Joules

Explanation / Answer

A) c for ice is 2090 j/kg B)latent heat of fusion of ice is 3..33*10^5 C)c of water is 4.19*10^3 D)latent heat of vapourisation of water is 2.26*10^6 WE HAVE TO MEMORISE THESE OR DATA MUST BE GIVEN>....!! PLEASE RATE IT...!!!

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