Question

$0.1{ m }^{ 3 }$ of water at ${ 80 }^{ \circ }C$ is mixed with $0.3{ m }^{ 3 }$ of water at${ 60 }^{ \circ }C$. The final temperature of the mixture is:
$A.\quad{ 65 }^{ \circ }C$
$B.\quad{ 70 }^{ \circ }C$
$C.\quad{ 60 }^{ \circ }C$
$D.\quad{ 75 }^{ \circ }C$

Answer 1

Hint: The first law of thermodynamics is also called law of conservation of energy. According to Law of conservation of energy, energy can neither be created nor be destroyed or ($\sum \ { Q= 0 }$). Hence hotter liquid will transfer its energy in the form of heat which will be absorbed by colder liquid so that net energy in the sample of two liquids is conserved.

Formula used: $Q=ms\Delta T$, $\rho = \dfrac mV$
Where $m$ = mass of sample to which heat is provided
$s$ = specific heat of the sample liquid
$\Delta T$ = Change in temperature on adding or ejecting heat.
$\rho$ = density of the sample liquid
$V$ = volume of sample liquid

Complete step-by-step answer:
As we know, total energy of the system should remain conserved, hence we can say that change in the total energy of the system must be zero, i.e. $\Delta { Q }_{ }=0$
Where ${ Q }$ is the total energy possessed by all the samples (two in this case).
& total energy $Q = { Q }_{ 1} +{ Q }_{ 2 }$
Hence we can write:
${ Q }_{ 1 }+{ Q }_{ 2 }=Constant$
Now, differentiating both sides, we get
${ Q }_{ 1 }+{ Q }_{ 2 }= 0$
and$Q=ms\Delta T$
so, ${ m }_{ 1 }s_{ 1 }\Delta T_{ 1 }+{ m }_{ 2 }s_{ 2 }\Delta T_{ 2 }=0$
Since both the samples are water,
Hence $s_{ 1 } = s_{ 2 } = s$ (say)
Putting the values of $\ {m_{1}}$ & ${m_{2}}$ in the equation, we get
$\Rightarrow$ $(\rho \times0.1)\times s\times[T-80^{ \circ }]+(\rho \times0.3)\times s\times[T-60^{ \circ }]=0$ {${ m }_{ 1 }=\rho .V_{ 1 }\quad and \quad m_2=\rho.V_2$ }
$\Rightarrow$ $0.1\times T-8+0.3\times T-18 = 0$
$\Rightarrow 0.4T=26^{ \circ }$
$\Rightarrow T=65^{ \circ }$

Additional information: Students might make a mistake where there’s a chance of transition change of the sample which happens for specific temperatures called boiling and freezing points. This concept will always work until transition changes.

Note: $\Delta T$ is the difference or change in temperature i.e. Final temperature – initial temperature. This is the most basic and important concept. If a student applies this concept in these types of questions, in which two or more, different or same liquids with different temperatures, are mixed and we are supposed to find the equilibrium temperature of the mixture, one must use this concept and there will be no place for any doubt.