Asked by rita
                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
            
            
        Answers
                    Answered by
            Damon
            
    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
    
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
                    Answered by
            mary
            
    Thanks
    
                    Answered by
            Favour
            
    What if the special heat capacity of water is 4.2×10^2
    
                    Answered by
            Shammah
            
    Where is the 4186 from
    
                    Answered by
            Opeyemi
            
    Yes
    
                    Answered by
            JENNIFER
            
    Where is the 4186 from? i am confused
    
                    Answered by
            That is not thd
            
    That not the answer
    
                    Answered by
            roger
            
    i dont even understand this solving sef
    
                    Answered by
            Faruq
            
    I don't understand the solving
    
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