# 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: $Q = m \cdot c \cdot \Delta T$ 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 $450 \setminus J k {g}^{-} 1 {K}^{-} 1$ and for water as $4200 \setminus J k {g}^{-} 1 {K}^{-} 1$)

Jun 2, 2018

Let final temperature of mixture $= {T}^{\circ} C$.

Heat lost by piece of iron ${Q}_{\text{lost}} = \frac{200}{1000} \times 450 \times \left(300 - T\right) \setminus J$

${Q}_{\text{lost}} = 90 \left(300 - T\right) \setminus J$

Heat gained water ${Q}_{\text{gained}} = 1.00 \times 4200 \times \left(T - 20\right) \setminus J$

${Q}_{\text{gained}} = 4200 \left(T - 20\right) \setminus J$

Using Law of Conservation of energy

${Q}_{\text{lost"=Q_"gained}}$

Inserting calculated values we get

$90 \left(300 - T\right) = 4200 \left(T - 20\right)$
$\implies 27000 - 90 T = 4200 T - 84000$
$\implies 4290 T = 27000 + 84000$
$\implies T = \frac{27000 + 84000}{4290}$
$\implies T = {25.9}^{\circ} C$, rounded to one decimal place

Jun 4, 2018

Let final temperature of mixture $= {T}^{\circ} C$.
We know that

Change in temperature $\Delta T = \left({T}_{\text{final"-T_"initial}}\right)$ and
Change in heat $\Delta Q = m s \Delta T$

$\therefore$Change in heat of iron $\Delta {Q}_{\text{iron}} = \frac{200}{1000} \times 450 \times \left(T - 300\right) \setminus J$

$\implies \Delta {Q}_{\text{iron}} = 90 \left(T - 300\right) \setminus J$

Change in heat of water $\Delta {Q}_{\text{water}} = 1.00 \times 4200 \times \left(T - 20\right) \setminus J$

$\implies \Delta {Q}_{\text{water}} = 4200 \left(T - 20\right) \setminus J$

Using Law of Conservation of energy

$\Delta {Q}_{\text{iron"+DeltaQ_"water}} = 0$

Inserting calculated values we get

$90 \left(T - 300\right) + 4200 \left(T - 20\right) = 0$
$\implies 90 T - 27000 + 4200 T - 84000 = 0$
$\implies 4290 T = 27000 + 84000$
$\implies T = \frac{27000 + 84000}{4290}$
$\implies T = {25.9}^{\circ} C$, rounded to one decimal place