-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.
-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?
i¡® ve got the first question... so can anyone help me with the second??
<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.
273k=how much in ¡ãC?
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To answer both questions, we need to calculate the total energy exchanged during various processes. Let's break down each step:
<1> Energy required to heat the ice from -32.8 ¡ãC to 0 ¡ãC:
Given:
Mass of ice (m) = 166.4 g = 0.1664 kg
Specific heat capacity of ice (Cice) = 2090 J/(kg ¡ãC)
Change in temperature (ΔT1) = 0 ¡ãC - (-32.8 ¡ãC) = 32.8 ¡ãC
The energy required is calculated using the formula: Q1 = m * Cice * ΔT1
Q1 = 0.1664 kg * 2090 J/(kg ¡ãC) * 32.8 ¡ãC
<2> Energy required to melt the ice into water at 0 ¡ãC:
Given:
Latent heat of fusion of water (Lf,water) = 3.33 × 10^5 J/kg
The energy required is calculated using the formula: Q2 = m * Lf,water
Q2 = 0.1664 kg * 3.33 × 10^5 J/kg
<3> Energy required to heat the melted water from 0 ¡ãC to 44.6 ¡ãC:
Given:
Change in temperature (ΔT2) = 44.6 ¡ãC - 0 ¡ãC = 44.6 ¡ãC
Specific heat capacity of water (Cwater) = 4186 J/(kg ¡ãC)
The energy required is calculated using the formula: Q3 = m * Cwater * ΔT2
Q3 = 0.1664 kg * 4186 J/(kg ¡ãC) * 44.6 ¡ãC
<4> Total energy supplied by steam:
The total energy supplied is the sum of Q1, Q2, and Q3: Q_total = Q1 + Q2 + Q3
Now, let's calculate the values:
Q1 = 0.1664 kg * 2090 J/(kg ¡ãC) * 32.8 ¡ãC
Q2 = 0.1664 kg * 3.33 × 10^5 J/kg
Q3 = 0.1664 kg * 4186 J/(kg ¡ãC) * 44.6 ¡ãC
Q_total = Q1 + Q2 + Q3
Once you have the value of Q_total, you will know the energy supplied by the steam to change the state of the ice to water at 44.6 ¡ãC.
<2> To find the final mass of water, we need to consider the mass of steam and the mass of melted ice:
Mass of steam = Mass of ice = 166.4 g = 0.1664 kg
The final mass of water will be the sum of the initial mass of steam and the mass of melted ice.
Final mass of water = Mass of steam + Mass of ice
Simplify and calculate the final mass of water.
Keep in mind that mass is conserved during phase changes, so the total mass at the end will be the same as the initial mass.
Hope this helps!