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