A liquid of specific heat s flows through a conducting tube in a bath, entering at 90°C and leaving at 50°C at 1 kg s⁻¹. A bath liquid of specific heat 2s enters at 20°C at 2 kg s⁻¹. At steady state the bath-liquid outlet temperature is 10x °C. Find x.
Physics · Mixed Chapter
JEE Mains

Question

A liquid of specific heat s flows through a conducting tube in a bath, entering at 90°C and leaving at 50°C at 1 kg s⁻¹. A bath liquid of specific heat 2s enters at 20°C at 2 kg s⁻¹. At steady state the bath-liquid outlet temperature is 10x °C. Find x.

Physics · Mixed Chapter

G1 Physics · Accepted Answer

Step 1 — Heat lost per second by tube liquid: P₁ = m₁s(90−50) = 1×s×40 = 40s. Step 2 — Heat gained per second by bath liquid: P₂ = m₂(2s)(T₂−20) = 2×2s(T₂−20). Step 3 — Steady-state heat balance: 40s=4s(T₂−20) ⇒ T₂−20=10. Thus T₂=30°C. Step 4 — Match 10x °C: 10x=30 ⇒ x=3. Answer: 3.