Q 11-10-118JEE MainJEE Main 2019 (11 Jan, Shift 2)Medium
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:
Answer: (C) $80^\circ$C
First mixture: $100s_A(10) = 50s_B(15) \Rightarrow s_A = 0.75s_B$.
Second mixture at temperature $T$:
$$100(0.75s_B)(100 - T) = 50s_B(T - 50)$$
$$7500 - 75T = 50T - 2500 \Rightarrow T = 80^\circ\text{C}$$
Solution by Sreeraj P, M.Sc Physics