What would be the final temperature if 50 g of H2O at 30degress celsius was adde

Question

What would be the final temperature if 50 g of H2O at 30degress celsius was added to 40 g of H2O at 90 degrees celsius?(Assume no loss of heat to the surroundings.) (285.9 J/g C)(50 g)(30 degrees celsius-Final Temp)=(285.9 J/gC)(40 G)(Final Temp-90 degrees celsius) What would be the final temperature if 50 g of H2O at 30degress celsius was added to 40 g of H2O at 90 degrees celsius?(Assume no loss of heat to the surroundings.) (285.9 J/g C)(50 g)(30 degrees celsius-Final Temp)=(285.9 J/gC)(40 G)(Final Temp-90 degrees celsius)

Explanation / Answer

         Let the finaltemperature be T         Heat lost by thehot water = 40 g * 4.184 J/g.oC * ( 90oC - T)          Heat gainedby the cold water = 50 g * 4.184J/g.oC * ( T - 30oC)              Sinceno heat is lost to the surroundings ,                Heatlost by the hot water =    Heat gained by thecold water                    40g * 4.184 J/g.oC * ( 90oC -T)   =   50 g * 4.184J/g.oC * ( T - 30oC)                        T = 56.66oC                       

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