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 heat#DeltaQ=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
