Recent content by trying_mybest

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    Equilibrium Temp of two fluids

    Thank you! I caught that mistake earlier, fixed it, and got the correct answer.
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    Equilibrium Temp of two fluids

    m1 = 5 gal * 3.79 L/gal * 1000 g/L = 18,750 g m2 = 50 gal * 3.79 L/gal * 1000 g/L = 187,500 g m1*C*(Tf - T1) = m2*C*(Tf - T2) m1CTf - m1CT1 = m2CTf - m2CT2 m1CTf - m2CTf = m1CT1 - m2CT2 Tf(m1 - m2) = m1T1 - m2T2 Tf = (m1T1 - m2T2) / (m1 - m2) Tf = (18,750*212 - 187,500*50) / (18,750 -...
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    Equilibrium Temp of two fluids

    Final temp should be ˚C. I used m1C∆T1 = m2C∆T2 I keep getting 0 for the final temp when it's supposed to be 18 C
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    Equilibrium Temp of two fluids

    Not really sure how to start this one.
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    Laten heat of fusion calculation

    My first attempt, the initial temp was 23.3, but then the second attempt was 25.6. I will make that correction on my calculation. Thank you so much for all the help!
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    Laten heat of fusion calculation

    This is what I think it should be: Qheat H2O lost = Qice to reach 0 C + Qmelt ice Qheat H2O lost = mH2O * Cv,H2O * ∆TH2O Qheat H2O lost = [0.25 L * 1000 g/1L] * (4.184 J/g*K) * (0-23.3) = -24,372 J Qice to reach 0 C = mice * Cv,ice * ∆Tice Qice to reach 0 C = (4 * 11) * (2.108 J/g*K) *...
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    Laten heat of fusion calculation

    This is the procedure. "Take a large glass or a bowl and fill it with 0.5 liter of water at room temperature, (or let it sit until it is at room temperature) Then estimate how many ice cubes (from your freezer) you think it will take to bring the glass or bowl with 1/2 liter of water down to 0...
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    Laten heat of fusion calculation

    Because I don't have a thermometer to measure it accurately, I'm making the assumption that the final temperature is 0˚ C since that's when the ice stopped melting. Also I redid the experiment. The initial temp of the water was 25.6 ˚C. I ended up using 11 ice cubes before the next one was...
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    Laten heat of fusion calculation

    The ice does the phase change I believe.
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    Laten heat of fusion calculation

    Q1 = m(ice) * Cv * ∆T = amount of heat the ice absorbs Q2 = m(H2O) * Lf = amount of heat the water loses Q1 = (4*8 g) * (2198 J/g*K) * (0-(-15)) = 1.055 * 10^6 J Q1=Q2 Q2 = 1.055 * 10^6 J = m(H2O) * Lf Lf = 1.055 * 10^6 J / m(H2O) = 1.055 * 10^6 J / (0.25 L * 1000 g/1L) = 4220 J/g My answer is...
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