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

Question

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=mcDeltaT#
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#)

Layla Ahmed · Accepted Answer

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

Leo Abeyta · Answer

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

Xavier Adame · Answer

undefined To find the final equilibrium temperature when a hot iron mass is placed in water, use the formula:

$m_1c_1(T_1) = m_2c_2(T_2)$

where:
- $m_1$ is the mass of the iron,
- $c_1$ is the specific heat capacity of iron,
- $T_1$ is the initial temperature of the iron,
- $m_2$ is the mass of water,
- $c_2$ is the specific heat capacity of water,
- $T_2$ is the initial temperature of the water.

Rearrange the formula to solve for the final equilibrium temperature ($T_f$):

$$T_f = \frac{m_1c_1T_1 + m_2c_2T_2}{m_1c_1 + m_2c_2}$$