10g of ice at 0c is mixed with 100g of water at 50c. What is the resultant temperature of mixture
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It requires 800cal to melt 10 g of ice giving 10g of water at 0C
those800cal came from 100g of water at 50C lowering its temperature 8degrees to 42C
So we have 10g at 0C and 100g at 42C
10x =100[42-x]; x = 420 - 10x; 11x=420; x=~38.2C
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those800cal came from 100g of water at 50C lowering its temperature 8degrees to 42C
So we have 10g at 0C and 100g at 42C
10x =100[42-x]; x = 420 - 10x; 11x=420; x=~38.2C
____________________
hope that helps!!
Please mark me as brainliest
sonisatyam278:
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