An aluminium vessel of mass 0.5 kg contains 0.2 kg of water at 20 degree Celsius. A block of iron of mass 0.2 kg at 100 degree celsius is gently put into water.

Question

An aluminium vessel of mass 0.5 kg contains 0.2 kg of water at 20 degree Celsius. A block of iron of mass 0.2 kg at 100 degree celsius is gently put into water. Find the equilibrium temperature of the mixture.

Given, specific heat of aluminium = 910 joule per kg per Kelvin

Specific heat of iron = 470 joule per kg per Kelvin

Specific heat of water = 4200 joule per kg per kelvin

Answer 1

I am assuming the aluminum vessel is at the same temperature as the water in the vessel.

 

The hot iron cools down and the water and aluminum warm up, with all 3 reaching an equilibrium temperature T_e somewhere between the initial temperature of the hot iron and the cool water and aluminum.

 

T_Fe = initial temperature of the hot iron

c_Fe = specific heat of iron 

m_Fe = mass of iron

 

T_Al = initial temperature of the cool aluminum

c_Al = specific heat of Al

m_Al = mass of the aluminum

 

T_H20 = initial temperature of the cool water = initial temperature of the cool aluminum

c_H2O = specific heat of the water

m_H2O = mass of the water

 

T_e = final equilibrium temperature

 

Heat flowing out of the hot iron as it cools down from T_Fe to T_e:

 

Q_Fe = m_Fec_Fe(T_Fe - T_e)

 

Heat flowing into the cool aluminum as it warms up from T_Al to T_e:

 

Q_Al = m_Alc_Al(T_e - T_Al)

 

Heat flowing into the cool water as it warms up from T_H2O to Te:

Q_H2O = m_H2Oc_H2O(Te - T_H2O)

 

From energy conservation,

 

Q_Fe = Q_Al + Q_H2O

 

Then

 

m_Fec_Fe(T_Fe - Te) = m_Alc_Al(Te - T_Al) + m_H2Oc_H2O(Te - T_H2O)

 

Everything in the equation above is given in the problem statement except T_e.  Just plug in the numbers and solve for T_e.  You should be able to finish from here.

Answer 2

Hi Adith!

Here are the relevant equations we need to figure out the answer to this question:

q=mcΔT

q₁ + q₂ + q₃ = 0

 

 

Where q is the total energy required to heat a substance, m is the mass of the substance, s is our specific heat constant (raise the temperature of 1 gram of substance by 1 degree celsius), and ΔT is our change in temperature and can be further defined as (T_f-T_i). 

We also know that temperature and energy are always conserved or are constant. The easiest constant to use (because we are not losing or gaining energy) is 0. So we can set all of our specific heat equations to 0.

 

So this means that we need to write a whole new formula to figure out the final temperature T_f. The formulas we use are the two expressed above.

 

After writing our new equation, let's pay attention to the values that we know for certain and the values we do not know for certain.

Hope this helps!