Q 11-10-035NEETJEE MainEAMCET 2000 (Engineering)Easy
Two liquids at temperatures $60°$C and $20°$C respectively have masses in the ratio $3 : 4$ and their specific heats in the ratio $4 : 5$. If the two liquids are mixed, the resultant temperature is
Answer: (D) $35°$C
Concept. Heat lost by the hot liquid $=$ heat gained by the cold one: $m_1c_1(60 - \theta) = m_2c_2(\theta - 20)$.
Step 1. Use the ratios. $m_1c_1 : m_2c_2 = (3 \times 4) : (4 \times 5) = 12 : 20$:
$$12(60 - \theta) = 20(\theta - 20)$$
Step 2. Solve. $720 + 400 = 32\theta \Rightarrow \theta = 35°$C.
Answer. $35°$C (Option D).
Solution by Sreeraj P, M.Sc Physics