Subtopic: Kinetics of microbial death – pasteurization and sterilization of milk and other liquid foods
Question Statement:
In parboiling, water:paddy ratio is 1.2. Water (cp=4.2 kJ kg⁻¹K⁻¹) is heated from 25°C to 85°C by condensation of steam (latent heat 2114 kJ kg⁻¹) via a tubular heat exchanger. 1 ton paddy at 30°C is poured into the hot water; mixture stabilizes at 75°C. Assuming no heat loss: P. steam supplied is 431 kg; Q. specific heat of paddy is 1.12 kJ kg⁻¹K⁻¹; R. steam supplied is 143 kg; S. specific heat of paddy is 2.11 kJ kg⁻¹K⁻¹
Options:
(A) P, Q
(B) Q, R
(C) R, S
(D) P, S
Official Answer Key:
B
Step-by-Step Solution & Explanation
Water mass = 1.2 × 1000 kg = 1200 kg. Steam heating water 25→85°C: Q_steam = 1200×4.2×(85−25) = 302,400 kJ. Steam mass = 302,400/2114 = 143.0 kg (statement R). Mixing hot water (85°C) with 1000 kg paddy (30°C) to stabilize at 75°C: heat lost by water = 1200×4.2×(85−75) = 50,400 kJ = heat gained by paddy = 1000×c_p×(75−30). c_p(paddy) = 50,400/(1000×45) = 1.12 kJ kg⁻¹K⁻¹ (statement Q). Correct pair: Q, R.
Practice This in Real CBT Exam Interface
Test yourself with the official TCS-iON palette, virtual calculator, and immediate performance analytics.