hello please help me as soon as possible if you can I am handicap student needin
ID: 1863550 • Letter: H
Question
hello please help me as soon as possible if you can I am handicap student needing help with this
24. (20 pts.) If you mix 9.5 g of 100 °C steam with 80 g of -12.0 °C ice, what percentage, if any, of the ice remains solid? If all of the ice melts, determine T of the system at thermal equilibrium. IMPORTANT: To get full credit on this problem you must clearly show your reasoning/state what quantities you are calculating! Just equations and numbers is not enough. (Cre-2090 J/kg-K, cwater = 4186 J/kg·K, L,-2.26 × 10s J/kg, Ly 3.33 x 10s J/kg)Explanation / Answer
let
m_ice = 80 g = 0.080 kg
m_steam = 9.5 g = 0.0095 kg
heat lost by the steam as it comes from 100 C steam to 0 degrees water,
Q_lost = m_steam*Lv + m_steam*C_water*(100 - 0)
= 0.0095*2.26*10^6 + 0.0095*4186*100
= 25446.7 J
Heat gained by ice at comes from -12 C to 0 C water,
Q_gained = m_ice*C_Ice*(0 - (-12)) + m_ice*Lf
= 0.08*2090*12 + 0.08*3.33*10^5
= 28646.4 J
Q_gained > Q_lost
so, total ice deos not melt.
let m is the mass of ice that melts.
now apply
heat lost by steam = heat gained by ice
Q_lost = Qgained
25446.7 = m_ice*C_Ice*(0 - (-12)) + m*Lf
25446.7 = 0.08*2090*12 + m*3.33*10^5
==> m = (25446.7 - 0.08*2090*12)/(3.33*10^5)
= 0.0704 kg
= 70.4 g
percentage of ice remains solid = (m_ice - m)*100/m_ice
= (80 - 70.4)*100/80
= 12% <<<<<<<<<------------------Answer
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