When 100 g of a liquid A at 100° C is added to 50 g of a liquid B at temperature 75° C, the temperature of the mixture becomes 90° C. The temperature of the mixture, if 100 g of liquid A at 100° C is…

Question

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

Accepted Answer

Correct answer: C. (c) JEE Main 2019 Heat loss = Heat gain = mS ∆ So, m_AS_A ∆ _A=m_BS_B ∆ _B ⇒100 S_A (100-90)=50 S_B (90-75) 2S_A=1.5S_B ⇒S_A= 34S_B Now, 1000 S_B (100- )=50 S_B ( -50) 2 ( 34 ) (100- )=( -50) 300-3 =2 -100 400=5 ⇒ =80°C

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

Solution

More Thermal Properties & Calorimetry PYQs