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