hello please help me as soon as possible if you can I am handicap student needin

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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