Question

A liquid $A$ of mass $100g$ at $100^\circ C$ is added to $50g$ of a liquid $B$ at temperature $75^\circ C,$ the temperature of the mixture becomes $90^\circ C$ . Now if $100g$ of liquid $A$ at $100^\circ C$ is added to $50g$ of liquid $B$ at $50^\circ C,$ temperature of the mixture will be

• A. $80^\circ C$
• B. $60^\circ C$
• C. $70^\circ C$
• D. $85^\circ C$

Answer 1

Answer: (A)

Heat loss will always be equal to heat gain.
As per $1^{s t}$ condition,
$100 \times S_{A} \times(100-90)=50 \times S_{B} \times(90-75)$
As per $2^{\text {nd }}$ condition,
$ 100 \times S_{ A }(100-\theta)=50 \times S_{ B }(\theta-50) $
Dividing (ii) by (i), we get
$ \frac{100-\theta}{100-90}=\frac{\theta-50}{90-75} $
$ 300-3 \theta=2 \theta-100 $
$ \theta=80^{\circ} C $