Heat and Thermodynamics 1 Question 3
3. When $100 g$ of a liquid $A$ at $100^{\circ} C$ is added to $50 g$ of a liquid $B$ at temperature $75^{\circ} C$, the temperature of the mixture becomes $90^{\circ} C$. The temperature of the mixture, if $100 g$ of liquid $A$ at $100^{\circ} C$ is added to $50 g$ of liquid $B$ at $50^{\circ} C$ will be
(a) $60^{\circ} C$
(b) $80^{\circ} C$
(c) $70^{\circ} C$
(d) $85^{\circ} C$
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Solution:
- In first case according to principle of calorimetry, heat lost by liquid $A=$ heat gained by liquid $B$
or
$$ m _A S _A \Delta T _A=m _B S _B \Delta T _B $$
where, $S _A$ is specific heat capacity of $A$ and $S _B$ is specific heat capacity of $B$
$$ \begin{aligned} & \Rightarrow \quad 100 \times S _A(100-90)=50 \times S _B(90-75) \\ & \Rightarrow \quad 1000 S _A=50 \times 15 S _B \\ & \text { or } \quad 4 S _A=3 S _B \end{aligned} $$
Similarly, in second case,
$$ 100 \times S _A(100-T)=50 \times S _B(T-50) $$
where, $T=$ Final temperature of the mixture.
$$ \Rightarrow \quad 4 S _A(100-T)=2 S _B(T-50) $$
Using Eq. (i),
$$ 3 S _B(100-T)=2 S _B(T-50) $$
$$ \begin{aligned} \text { or } & 300-3 T & =2 T-100 \\ \text { or } & 5 T & =400 \\ \text { or } & T & =80^{\circ} C \end{aligned} $$
