-Steam at 100 ¡ãC is mixed with 166.4 g of ice at ¨C32.8 ¡ãC, in a thermally insulated container, to produce water at 44.6 ¡ãC. Ignore any heat absorption by the container.

Question

-Cwater = 4186. J/(kg ¡ãC)
-Cice = 2090. J/(kg ¡ãC)
-Lf,water = 3.33 ¡Á 105 J/kg
-Lv,water = 2.26 ¡Á 106 J/kg
Also,energy is required to bring all the ice up to 0 ¡ãC=11400J; energy is required to melt the ice into water at 0 ¡ãC=55400J; energy required to raise the temperature of this melted water to 44.6 ¡ãC =31100J;

<1>How much energy must then be supplied by the steam to change the state of 166.4 g of ice at ¨C32.8 ¡ãC to water at 44.6 ¡ãC?

<2>hat is the final mass of water in the cup at 44.6 ¡ãC?

Answer 1

i¡® ve got the first question... so can anyone help me with the second??

Answer 2

<1> Add all the previous energy together: 11400 + 55400 + 31100J (as per question).

<2> Q is from <1>.
t.f. <1> = m(steam)Lv + m(steam)c(water) * (100 - T(water)*C)
Convert that answer to g, because you get it in kg.
That answer, plus the initial mass of 166.4g, is your total mass of water in the cup.

Answer 3

273k=how much in ¡ãC?

Answer 4

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