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Math Help - Heat problems

  1. #1
    Junior Member ihmth's Avatar
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    Heat problems

    1) Determine the amount of heat needed to turn 500g of ice in a 100g aluminum pan into steam.

    2) What will be the phase of 1 liter of water in a 200g copper pot if it is placed in a coolant that absorbs 100,000J of heat?

    3) How much heat is required to turn 50g of ice with temperature of -50 C to steam at 150 C?

    Specific Heat:
    Water = 4180 J/kg*K or 1.000 cal/g*C
    Copper = 392 J/kg*K or 0.093 cal/g*C
    Aluminum = 895 J/kg*K or 0.214 cal/g*C


    Latent Heat:

    = Heat of fusion
    Water = 3.34 * 10^5 J/kg or 80 cal/g
    Copper = 2.05 * 10^5 J/kg or 48.9 cal/g
    Aluminum = 3.76 * 10^5 J/kg or 89.9 cal/g

    = Heat of vaporization
    Water = 22.6 * 10^5 J/kg or 540 cal/g
    Copper = 48 * 10^5 J/kg or 1150 cal/g
    Aluminum = 113.71 * 10^5 J/kg or 2717.7 cal/g
    Last edited by ihmth; February 24th 2008 at 11:55 PM.
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by ihmth View Post
    1) Determine the amount of heat needed to turn 500g of ice in a 100g aluminum pan into steam. My answer is 360,000 cal
    If you are going to specify that we have an aluminum pan then we need to know what temperature the pan is starting at. Additionally we need to know what temperature the ice starts at as well. There is not enough information present to do this problem.

    -Dan
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  3. #3
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by ihmth View Post
    2) What will be the phase of 1 liter of water in a 200g copper pot if it is placed in a coolant that absorbs 100,000J of heat? I dont have an answer

    Specific Heat:
    Water = 4180 J/kg*K or 1.000 cal/g*C
    Copper = 392 J/kg*K or 0.093 cal/g*C

    Latent Heat:

    = Heat of fusion
    Water = 3.34 * 10^5 J/kg or 80 cal/g
    The maximum amount of heat energy the water can have is when its initial temperature is at the boiling point, 373 K. (Otherwise it is steam, not water.) Assume the copper pot also has this temperature at the beginning.

    Let's find out how much heat needs to be absorbed from the system to turn the water at the boiling point into ice:
    c_wm_w(373 - 273) + c_{Cu}m_{Cu}(373 - 273) + L_wm_w = 7.5984 \times 10^5~J
    (The copper pot must be a maximum of 273 K for the water to freeze, so I'm assuming that it drops at the same temperature rate as the water. This is not quite physically true, but will do as an approximation.) The first two terms take care of the heat needed to be absorbed to take the water and copper from 100 C to 0 C, the third term is the heat to be absorbed from the water at 0 C to turn it into ice.

    The coolant absorbs 1 \times 10^5~J of heat energy. This is less than the heat we can absorb from the water and copper at 100 C to freeze the ice, so we would need a starting temperature for the water-ice system in order to give a definitive answer to this question.

    (Now if the coolant absorbed, say, 1 \times 10^6~J then we would know for sure that the water would freeze.)

    -Dan
    Last edited by topsquark; February 23rd 2008 at 10:59 AM.
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by ihmth View Post
    3) How much heat is required to turn 50g of ice with temperature of -50 C to steam at 150 C? My answer is 41,000 cal

    Specific Heat:
    Water = 4180 J/kg*K or 1.000 cal/g*C

    Latent Heat of fusion:
    Water = 3.34 * 10^5 J/kg or 80 cal/g

    = Heat of vaporization
    Water = 22.6 * 10^5 J/kg or 540 cal/g
    This one we can do:
    Heat needed to turn the 50 g of ice at -50 C to ice at 0 C:
    c_im_i(0 - -50) = 5125~J
    ( c_i = 2.050~J/g ^o C. You didn't give this, I had to look it up. It's Wikipedia so I'm only assuming the numbers are correct.)

    Heat needed to turn 50 g of ice at 0 C to water at 0 C:
    L_fm_i = 16700~J

    Heat needed to turn 50 g water at 0 C to water at 100 C:
    c_wm_w(100 - 0) = 20900~J

    Heat needed to turn 50 g water at 100 C to steam at 100 C:
    L_vm_w = 113000~J

    Finally, heat need to turn 50 g of steam at 100 C to steam at 150 C:
    c_sm_s(150 - 100) = 5200~J
    ( c_s = 2.080~J/g ^oC. Again, you didn't give us this information.)

    Adding these up gives a total of 160925 J of heat required.

    In the future, please provide us with all the information you are given.

    -Dan
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  5. #5
    Junior Member ihmth's Avatar
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    Quote Originally Posted by topsquark View Post
    If you are going to specify that we have an aluminum pan then we need to know what temperature the pan is starting at. Additionally we need to know what temperature the ice starts at as well. There is not enough information present to do this problem.

    -Dan
    ice and the aluminum pan is at 0 C
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  6. #6
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by ihmth View Post
    ice and the aluminum pan is at 0 C
    Quote Originally Posted by ihmth View Post
    1) Determine the amount of heat needed to turn 500g of ice in a 100g aluminum pan into steam. My answer is 360,000 cal

    Specific Heat:
    Water = 4180 J/kg*K or 1.000 cal/g*C
    Aluminum = 895 J/kg*K or 0.214 cal/g*C


    Latent Heat:

    = Heat of fusion
    Water = 3.34 * 10^5 J/kg or 80 cal/g

    = Heat of vaporization
    Water = 22.6 * 10^5 J/kg or 540 cal/g
    So
    L_f m_i + c_wm_i(100 - 0) + c_{Al}m_{Al}(100 - 0) + L_vm_i

    -Dan
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