Answer:
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
Answer:
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
