If a 100.0 g sample of water at 6.7°C is added to a 100.0 g sample of water at 57.0°C, determine the final temperature of the water. Assume no heat is lost to the surroundings.

The answer to your question is  Te = 31.85 °C

Explanation:

Data

mass 1 = 100 g

temperature 1 = 6.7°C

mass 2 = 100 g

temperature 2 = 57°C

equilibrium temperature = ?

C = 1

Formula

                  mCΔT = - mCΔT

Substitution

                  100(Te - 6.7) = - 100(Te - 57)

Simplification

                  100 Te - 670 = - 100 Te + 5700

                  100Te + 100Te = 5700 + 670

                                 200Te = 6370

                                         Te = 6370 / 200

Result

                                          Te = 31.85 °C

The final temperature will be 31.85 °C

Explanation:

Step 1: Data given

Mass of water = 100.0 grams

Temperature of water = 6.7 °C

Mass of the other sample water = 100.0 grams

Temperature = 57.0 °C

Step 2: Calculate the final temperature

Energy Lost by the hot water = Energy Gain by the cold water

mass of hot water*c*(Temp. of hot water-x) = mass of cold water*c*(x-Temp. of cold water)

Energy Lost by the hot water=Energy Gain by the cold water

mass of hot water*c*(Temp. of hot water-x)=mass of cold water*c*(x-Temp. of cold water)

100*c*(57-x)=100*c*(x-6.7)

100*(57-x)=100*(x-6.7)

5700 - 100x = 100x -670

6370 = 200x

x = 31.85

The final temperature will be 31.85 °C

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