# How to find the final equilibrium temperature, when a hot iron mass is placed in water?

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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