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When 167.2 joules of heat is added to 4.00 grams of water at 10c the resulting temperature is what

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Q=mc \Delta T \\\\ Q=168,2J\\ m=4g\\ c=4.184(J)/(g^(o)C)\\\\ \Delta T=(Q)/(mc)=(167,2J)/(4g*4,184(J)/(g^(o)C))=9,99^(o)C\\\\ \Delta T = T_(f)-T_(i)\\\\ 9,99^(o)C=T_(f)-10^(o)C\\\\ T_(f)=9,99^(o)C+10^(o)C=19,99^(o)C\approx20^(o)C
User Lorenzo Pichilli
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