JEE PYQ: Thermodynamics Question 42
Question 42 - 2019 (11 Jan Shift 2)
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:
(1) 85$°$C
(2) 60$°$C
(3) 80$°$C
(4) 70$°$C
Show Answer
Answer: (3)
Solution
Heat loss = Heat gain
$m_A S_A \Delta\theta_A = m_B S_B \Delta\theta_B$
$100 \times S_A \times (100 - 90) = 50 \times S_B \times (90 - 75)$
$2S_A = 1.5S_B \Rightarrow S_A = \frac{3}{4}S_B$
Now, $100 \times S_A \times (100 - \theta) = 50 \times S_B \times (\theta - 50)$
$2 \times \frac{3}{4} \times (100 - \theta) = (\theta - 50)$
$300 - 30 = 20 - 100$
$400 = 50 \Rightarrow \theta = 80°$C
BETA
