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A 42.14−g sample of water at 87.8°C is added to a sample of water at 25.4°C...

A 42.14−g sample of water at 87.8°C is added to a sample of water at 25.4°C in a constant-pressure calorimeter. If the final temperature of the combined water is 40.1°C and the heat capacity of the calorimeter is 26.3 J/°C, calculate the mass of the water originally in the calorimeter. Enter your answer in scientific notation.

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Answer #1

-qH2O(hot)     = qH2O + q calorimeter

-mcT    = mcT + cp*T

-42.14*4.184*(40.1-87.8)   = m*4.184*(40.1-25.4) + 26.3*(40.1-25.4)

8410.16                            = m*61.5048 + 386.61

m    = 130.5g = 1.305*10^2 g

The mass of water is 130.5g = 1.305*10^2 >>>>answer

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