Question

When 100 g of a liquid A at 100^oC is added to 50 g of a liquid B at temperature 75^o C , the temperature of the mixture becomes 90^oC. The temperature of the mixture, if 100 g of liquid A at 100^oC is added to 50 g of liquid B at 50^oC, will be:
A. 80^oC
B 60^oC
C. 70^oC
D 85^oC

Answer 1

Hint: As we all know that when two bodies at different temperatures are kept in contact with each other, the heat transfer takes place through a body of higher heat potential to the body at lower heat potential.

Complete step by step answer:
Here we can apply the principle of calorimetry i.e. the body at lower temperature would absorb heat and body at higher temperature would reject heat. So the total heat gained by the body is equal to the heat lost by the other body. So, we can write the equation as,
$MC\Delta T = mc\Delta t$
$MC\left( {100 - {T_1}} \right) = mc(T - {t_1})$ …… (I)
Here M is the mass of liquid A, C is the specific heat of liquid A, ${T_1}$ is the initial temperature of liquid A, m is the mass of liquid B, ${t_1}$ is the initial temperature of liquid B, $c$ is the specific heat of liquid B and $T$ is the final temperature.
We can now substitute, $M = 100\,{\text{g}}$ , $m = 50\,{\text{g}}$ ${T_1} = 100^\circ {\text{C}}$ , ${t_1} = 75^\circ {\text{C}}$ , $T = 90^\circ {\text{C}}$ , in the above relation to find the relation between $C$ and $c$.
$
   \Rightarrow 100\;{\text{g}} \times {\text{C}} \times {\text{(100 - 90) = 50}} \times c \times (90 - 75) \\
   \Rightarrow 2 \times C \times 10{\text{ = c}} \times 15 \\
 $
$ \Rightarrow C = \dfrac{3}{4}c$ …… (II)
Now coming here for the second case, we can now substitute $M = 100\;{\text{g}}$, , $m = 50\,{\text{g}}$ ${T_1} = 100^\circ {\text{C}}$ , ${t_1} = 50^\circ {\text{C}}$ , to find the value of $T$ in equation (I), we will get.
$ \Rightarrow 100\,{\text{g}} \times {\text{C}} \times (100^\circ {\text{C - }}{{\text{T}}_f}) = 50 \times c \times ({{\text{T}}_f} - 50\;^\circ {\text{C}})$
We will now substitute $C = \dfrac{3}{4}c$ here to find the value of final temperature ${T_f}$ .
$
   \Rightarrow 100\,{\text{g}} \times \dfrac{3}{4}c \times (100^\circ {\text{C - }}{{\text{T}}_f}) = 50 \times c \times ({{\text{T}}_f} - 50\;^\circ {\text{C}}) \\
   \Rightarrow 25 \times 3 \times (100{\text{ - }}{{\text{T}}_f}) = 50 \times ({{\text{T}}_f} - 50\;) \\
   \Rightarrow 3 \times (100{\text{ - }}{{\text{T}}_f}) = 2 \times ({{\text{T}}_f} - 50\;) \\
   \Rightarrow 400 = 5{{\text{T}}_f} \\
  \therefore {{\text{T}}_f} = 80\;^\circ {\text{C}} \\
 $
Therefore, the final temperature of the mixture is $80\;^\circ {\text{C}}$.

Note: We must keep in mind the principle of calorimetry is generally the basic energy conservation. It means that the amount of energy gained by one system is the amount of energy lost by another system if the bodies are exchanging energy.