Thermal Physics Problem -- Dropping a hot mass of iron into 20C water

[Woopa]

Homework Statement

A piece of iron of mass 200g and temperature 300 degrees celcius is dropped into 1.00kg of water of temperature 20 degrees celcius.

Predict the final equilibrium of the water.

(Take c for iron as 540 J/kg/K and for water as 4200 J/kg/K).

Relevant Equations

Q=mcΔT)

Hi,

The solution for this question is

thermal energy lost= thermal energy gained
0.200x450x(300-T)=1.0x4200x(T-20)
T=26 degrees celsius.

However, I am struggling to grasp why (300-T) is used.

I have always known a change in something to be final - initial. Therefore change in T= Final- initial.

However, in the case of (300-T) Initial- Final is being used for the change and I am struggling to understand why.

 

[Lnewqban]

The temperature of the piece of iron has been reduced from the initial 300 to the final common value of T, which is shared by iron and water once the heat transfer has ended.
Please, note a typo error (540 versus 450).
You can read more about thermal equilibrium here:
https://www.physicsclassroom.com/Class/thermalP/u18l1d.cfm

 

[kuruman]

The solution is poorly presented and that is why you are struggling to understand it. The starting equation is
$m_1 c_1 \Delta T_1 + m_2 c_2\Delta T_2=0$.
If you replace the Deltas with actual differences, you get
$m_1 c_1 (T-T_1) + m_2 c_2(T-T_2)=0$.
Now move one of the terms to the other side of the equation and change sign
$m_1 c_1 (T-T_1) = - m_2 c_2(T-T_2) =+ m_2 c_2(T_2-T).$

 

[Woopa]

The solution is poorly presented and that is why you are struggling to understand it. The starting equation is
$m_1 c_1 \Delta T_1 + m_2 c_2\Delta T_2=0$.
If you replace the Deltas with actual differences, you get
$m_1 c_1 (T-T_1) + m_2 c_2(T-T_2)=0$.
Now move one of the terms to the other side of the equation and change sign
$m_1 c_1 (T-T_1) = - m_2 c_2(T-T_2) =+ m_2 c_2(T_2-T).$

Excellent I've got it! Thank you!

 

[kuruman]

Excellent I've got it! Thank you!

You are welcome. BTW if you use the heat capacity $C$ instead of the product of mass and specific heat heat, the answer takes a familiar form $$T=\frac{C_1T_1+C_2T_2}{C_1+C_2}$$ which is just like the expression for the center of mass of two masses. The final equilibrium temperature is the heat-capacity-averaged temperature much like the center of mass is the mass-averaged position. With this in mind, you can have $N$ objects at different initial temperatures thrown together in a calorimeter and be able to write down the final temperature immediately:$$T_{\text{final}}=\frac{m_1c_1T_1+m_2c_2T_2+\dots+m_Nc_NT_N}{m_1c_1+m_2c_2+\dots+m_Nc_N}.$$