How to find the final equilibrium temperature, when a hot iron mass is placed in water?
This is the question in my textbook. I checked the answers but did not understand how they got to their answer. I used the following equation:
Q=m*c*DeltaT
The textbook answer is 25.9°C.
Question:
A piece of iron of mass 200 g and temperature 300 °C is dropped into 1.00 kg of water of temperature 20 °C. Predict the final equilibrium temperature of the water. (Take c for iron as 450\ Jkg^-1K^-1 and for water as 4200\ Jkg^-1K^-1 )
This is the question in my textbook. I checked the answers but did not understand how they got to their answer. I used the following equation:
The textbook answer is 25.9°C.
Question:
A piece of iron of mass 200 g and temperature 300 °C is dropped into 1.00 kg of water of temperature 20 °C. Predict the final equilibrium temperature of the water. (Take c for iron as
2 Answers
Let final temperature of mixture
Heat lost by piece of iron
Q_"lost"=90(300-T)\ J
Heat gained water
Q_"gained"=4200(T-20)\ J
Using Law of Conservation of energy
Q_"lost"=Q_"gained"
Inserting calculated values we get
90(300-T)=4200(T-20)
=>27000-90T=4200T-84000
=>4290T=27000+84000
=>T=(27000+84000)/4290
=>T=25.9^@C , rounded to one decimal place
Let final temperature of mixture
We know that
Change in temperature
DeltaT=(T_"final"-T_"initial") and
Change in heatDeltaQ=msDeltaT
=>DeltaQ_"iron"=90(T-300)\ J
Change in heat of water
=>DeltaQ_"water"=4200(T-20)\ J
Using Law of Conservation of energy
DeltaQ_"iron"+DeltaQ_"water"=0
Inserting calculated values we get
90(T-300)+4200(T-20)=0
=>90T-27000+4200T-84000=0
=>4290T=27000+84000
=>T=(27000+84000)/4290
=>T=25.9^@C , rounded to one decimal place