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    whatevaxd's Avatar
    whatevaxd Posts: 26, Reputation: 1
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    #1

    Jul 24, 2009, 05:56 PM
    Physics Phase Changes
    1.A heater can supply heat at the rate of 4.8x10^2 J s-1. How long will it take to boil away 12.5g of water originally at 26.5 Celsius?

    For this question, I am not sure whether do I need to use the formula Q=mL.
    I used Q=mcT and I got 3840.38 J.

    2.2.15kg of water at 21.5 Celsius into the freezer. It takes 2 hours to freeze all of the water, calculate the rate the freezer is removing the heat.

    For this question, I am also unsure whether do I need to use the formula Q=mL.

    3.In hot conditions, a person will sweat between 1.5L and 4L of water every hour.

    a.If a person evaporates 4L of water from their sweat, how much heat enrgy does the person lose. Assume the evaporating sweat does not get heat from anywhere else. Normal human body core temperature is 37 Celsius.

    b.By how much would a girl's temperature rise if the 4L of sweat did not evaporate from her skin? Her mass is 55 kg. Assume her body produces heat at a constant rate and she loses heat in no other way.

    I am not sure how to do question number 3. Some explaining would be awesome. Thx
    Unknown008's Avatar
    Unknown008 Posts: 8,076, Reputation: 723
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    #2

    Jul 26, 2009, 02:58 AM
    Quote Originally Posted by whatevaxd View Post
    1.A heater can supply heat at the rate of 4.8x10^2 J s-1. How long will it take to boil away 12.5g of water originally at 26.5 Celsius?

    For this question, i am not sure whether do i need to use the formula Q=mL.
    I used Q=mcT and i got 3840.38 J.
    You need both. You need Q=mcT for raising the tem[perature of the water from 26.5 C to 100 C and use Q=mL for boiling water away.

    2.2.15kg of water at 21.5 Celsius into the freezer. It takes 2 hours to freeze all of the water, calculate the rate the freezer is removing the heat.

    For this question, i am also unsure whether do i need to use the formula Q=mL.
    Here also, use Q=mcT for the decrease from 21.5 C to 0 C, then Q=mL to freeze the water.

    3.In hot conditions, a person will sweat between 1.5L and 4L of water every hour.

    a.If a person evaporates 4L of water from their sweat, how much heat enrgy does the person lose. Assume the evaporating sweat does not get heat from anywhere else. Normal human body core temperature is 37 Celsius.

    b.By how much would a girl's temperature rise if the 4L of sweat did not evaporate from her skin? Her mass is 55 kg. Assume her body produces heat at a constant rate and she loses heat in no other way.

    I am not sure how to do question number 3. Some explaining would be awesome. Thx
    a. Use Q=mL.
    b. Here, you have to assume that the heat you previously found is the excess heat. That means, that is also the amount of heat produced in an hour. Use Q=mcT to find her temperature rise. You just found Q, you know m (55kg) and c is a constant. T is what you're looking for.



    When you are dealing with a change in state, you use Q = mL
    When you are dealing with a change in temperature only, you use Q = mcT.
    whatevaxd's Avatar
    whatevaxd Posts: 26, Reputation: 1
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    #3

    Jul 27, 2009, 12:39 AM

    I managed to get all the other questions except for question 3, can u give me a hand?

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