Anonymous asked in Science & MathematicsPhysics · 7 years ago

If 50.0 g of ice at 0 C is added to 150.0 g of water at 80.0 C?

What will the temperature of the mixture be when it reaches thermal equilibrium?

3 Answers

  • 7 years ago
    Favorite Answer

    mass of ice (m1) = 50 grams

    mass of water (m2) = 150 grams

    Temperature of water (T2)= 80.0 C

    Temp. of ice (T1) = 0 C

    Temperature of mixture (Tm)= ?

    heat of melting ice(L) = 80 cal/g

    heat capacity of water (c) = 1 cal/g

    m1.L + m1.c.ΔTice = m2.c.ΔTwater

    m1.L + m1.c.(Tm-T1) = m2.c.(T2-Tm)

    50.80 + 50.1(Tm-0) = 150.1.(80-Tm)

    4000 + 50Tm = 12000 -150Tm

    200 Tm = 8000

    Tm = 40 C

    ok? rate please

  • 7 years ago

    I cannot do the math. I encountered a roadblock in my math that I cannot overcome. However, I can give you and any future people answering this question this tip:

    You will need to find the total amount of energy transferred and include the heat of fusion of ice. Here is what I can help with.

    The amount of energy transferred into the ice to melt it: Q=m*Hf 16700 J = .050*(3.34*10^5)

    After considering the melted ice, we can now find the final temperature by using this formula:

    The water we gained from melting the ice will be represented by "i", and temperature will be measured in Kelvins

    Final Temperature = ((mass of i)*Ci*(initial temp)+(mass of w)*Cw*(initial temp)/((mass of i)*Ci+(mass of w)*Cw)

    Final Temperature = (.050*4180*273+.150*4180*(initial temp))/(.050*4180+.150*4180)

    Final Temperature = (57057+627*(initial temp))/(836)

    The only variable I am missing is the temperature of the .150 kg of water after melting the ice.

    Source(s): Physics class - Thermodynamics
  • 7 years ago

    (50.0 * 0 + 150.0 * 80)/(50+150) = 60 C

    This is assuming its a perfect system.

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