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Quick Physics Question

Magus`Magus` The fun has been DOUBLED!Registered User regular
edited November 2007 in Help / Advice Forum
It's a simple question and I know the answer, but I'm having trouble showing the work (which is basically what I'm being graded on).

Alright, we have 300g of water at 50 degrees Celsius. Next we have 200g of water at 100 degrees Celsius.

Now, if they were equal amounts of water and had the same temps, then we we mixed we'd hit exactly half way (75 degrees Celsius).

Since we have more of one than the other, it will NOT be half. Also, since a majority of the water started at the lower temp, the final temp will be less than 75 degrees, the halfway point. However, this is kind of where I've hit a stall.

I've been trying the various equations I've been taught and I can't seem to get the given answer (70 degrees).

The teacher wrote the following things as 'hints'. One is heat gained + heat lost = 0. I knew this one as it's due to the conservation of energy. I'm still trying to figure out what this hint is supposed to be telling me to no avail.

The second equation that I've given as a hint is Q = mass x calories x change in temp. I'm even more lost by this. There is a similar puzzle in the book but it talks about comparing Calories (big C!) between Ethanol and Water.

So if anyone can kind of nudge me in the right direction, I'd appreciate it.

Magus` on

Posts

  • BoutrosBoutros Registered User regular
    edited November 2007
    Magus` wrote: »
    It's a simple question and I know the answer, but I'm having trouble showing the work (which is basically what I'm being graded on).

    Alright, we have 300g of water at 50 degrees Celsius. Next we have 200g of water at 100 degrees Celsius.

    Now, if they were equal amounts of water and had the same temps, then we we mixed we'd hit exactly half way (75 degrees Celsius).

    Since we have more of one than the other, it will NOT be half. Also, since a majority of the water started at the lower temp, the final temp will be less than 75 degrees, the halfway point. However, this is kind of where I've hit a stall.

    I've been trying the various equations I've been taught and I can't seem to get the given answer (70 degrees).

    The teacher wrote the following things as 'hints'. One is heat gained + heat lost = 0. I knew this one as it's due to the conservation of energy. I'm still trying to figure out what this hint is supposed to be telling me to no avail.

    The second equation that I've given as a hint is Q = mass x calories x change in temp. I'm even more lost by this. There is a similar puzzle in the book but it talks about comparing Calories (big C!) between Ethanol and Water.

    So if anyone can kind of nudge me in the right direction, I'd appreciate it.

    The heat gained by the cold water is equal to the heat lost by the hot water so the total heat stays the same, so Q1 + Q2 = Q3 where Q1 is the heat of the warm water, Q2 is the heat of the cold water, and Q3 is the heat of the 500 grams of water at equilibrium. So you know all of the masses, all but one of the temperatures, and you know C, which cancels anyway, so all you have to do is solve for the temperature you don't know.

    Boutros on
  • PeekingDuckPeekingDuck __BANNED USERS regular
    edited November 2007
    Since you know that heat gained + heat lost = 0... you can substitute in Q (heat as energy) into your equation for your two liquids.

    Giving... m1(c1)(Final Temp - Initial Temp1) + m2(c2)(Final Temp - Initial Temp2) = 0

    I think your c (c1 and c2 are the same in this case) letter is referring to the specific heat of water... not calories.

    Where the 1 and 2 denote the 1st and 2nd liquids... this should help you solve it.

    PeekingDuck on
  • PeekingDuckPeekingDuck __BANNED USERS regular
    edited November 2007
    Well c does refer to calories since the units of c can be kcal/kg*deg C... but that's really not important in this problem.

    PeekingDuck on
  • LogicowLogicow Registered User regular
    edited November 2007
    Alright so you have a population of seagulls.

    200 seagulls are 100 years old, and 300 seagulls are 50 years old.

    What is the average age of a seagull in this population?

    200 seagulls * 100 years + 300 seagulls * 50 years = 35000 years total.

    3500 years total, divided among 500 seagulls, means 70 years per seagull. That's an average.

    Logicow on
  • Al_watAl_wat Registered User regular
    edited November 2007
    Logicow wrote: »
    Alright so you have a population of seagulls.

    200 seagulls are 100 years old, and 300 seagulls are 50 years old.

    What is the average age of a seagull in this population?

    200 seagulls * 100 years + 300 seagulls * 50 years = 35000 years total.

    3500 years total, divided among 500 seagulls, means 70 years per seagull. That's an average.

    while that would work and that is how i was thinking of the answer, hes gonna need to do it using physics equations.

    Al_wat on
  • PeekingDuckPeekingDuck __BANNED USERS regular
    edited November 2007
    That only works if both birds are seagulls. ;)

    PeekingDuck on
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