Question

When 100 g of a liquid A at ${100}^0$C is added to 50 g of a liquid B at temperature ${75}^0$C, the temperature of the mixture becomes ${90}^0$C. The temperature of the mixture, if 100 g of liquid A at ${100}^0$C is added to 50 g of liquid B at ${50}^0$C, will be:

• A. ${60}^0$C
• B. ${70}^0$C
• C. ${85}^0$C
• D. ${80}^0$C

Answer 1

The correct option is D ${80}^0$C

To find final temperature of mixture:

$100\times S_A\times [100-90]=50\times S_B\times (90-75)$

$2S_A=1.5S_B$

$S_A=\frac{3}{4}{S}_B$

Now, $100\times S_A\times [100-T]=50\times S_B(T=50)$

$2\times \left(\frac{3}{4}\right)(100-T)=(T-50)$

$300-3T=2T-100$

$400=5T$

$T=80$

Hence option D is correct