A piece of copper of mass 300 g at a temperature of 950 degree Celsius is quickly transferred to a vessel of negligible thermal containing 250 g of water at 25 degree Celsius. If the final temperature of the mixture is 100 degree Celsius calculate the mass of the water that will boil away ( specific heat capacity of copper =400 j/kg.k specific latent heat of vaporization of steam =2260000jkg

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  1. 0.300 * 400 * (950-100) = 102,000 Joules lost by copper

    that heats 0.250 kg water to 100 from 25
    0.250 * 4186 * 75 = 78488 Joules used from copper to heat water.
    That leaves 23,512 Joules to boil water
    23,512 = m 2,260,000
    m = .0104 kg = 10.4 grams

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  2. Thanks

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  3. What if the special heat capacity of water is 4.2×10^2

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  4. Where is the 4186 from

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  5. Yes

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  6. Where is the 4186 from? i am confused

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  7. That not the answer

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