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#)

2 Answers
Jun 2, 2018

Let final temperature of mixture #=T^@C#.

Heat lost by piece of iron #Q_"lost"=200/1000xx450xx(300-T)\ J#

#Q_"lost"=90(300-T)\ J#

Heat gained water #Q_"gained"=1.00xx4200xx(T-20)\ J#

#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

Jun 4, 2018

Let final temperature of mixture #=T^@C#.
We know that

Change in temperature #DeltaT=(T_"final"-T_"initial")# and
Change in heat #DeltaQ=msDeltaT#

#:.#Change in heat of iron #DeltaQ_"iron"=200/1000xx450xx(T-300)\ J#

#=>DeltaQ_"iron"=90(T-300)\ J#

Change in heat of water #DeltaQ_"water"=1.00xx4200xx(T-20)\ J#

#=>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