Question

What mass of steam at 100∘C must be added to 1.00 kg of ice at 0∘C to yield liquid water at 17 ∘C? The heat of fusion for water is 333 kJ/kg , the specific heat is 4186 J/kg⋅C∘ , the heat of vaporization is 2260 kJ/kg .

Express your answer to two significant figures and include the appropriate units.

Answer 1

eaparature of Steam, t= 100 100 °C mass of ice, mi = lng ice to initial temp. of ice, to = 0°c. 2. heat required Ikg errature 0° luga water of temp H = mil o.a 1x3338103 5 required to become is, = 333x103 J. Given, L=333 kilug ooc Water 2. heat heat required to become they to lug 17°C is, 1 0 Hg msht 1X4186 X 17 71162 J. Cas s = 41835/eg m.ng 2 et mass of steama then, heat rejected by stream to become from 100°C water 100°c vapour is. Hi (as heat of 3 J. mx 2260 X 10 reporisation = 2260 x103 kr/4. my mug mug water of 100°c heat r. rejected Hamsat 347438 m to become 19°c ug walkis, mx 418 x floor) Al

HO Then, for thermal equilibrium, HipHe'= AITAL 347438 Morava + 2260 x103) m = (333x103)+71162 a you 162 2607438 m m ug ~ 0.16 uge 2. ma 0.1550 0.16 16 ug Steam must be added. CB