Question

In a cold storage plant, 5000 kg potato having a constant specific heat capacity of 3.65 kJ kg\(^{-1}\) °C\(^{-1}\) are cooled from 28°C to 2°C in 24 hours. The heat of respiration of potato per 24 hour is 3.12 kJ kg\(^{-1}\) during the storage. Assuming the efficiency of the storage plant to be 70%, the capacity of the plant in ton of refrigeration (round off to 2 decimal places) is \(\underline{\hspace{2cm}}\).

Hint:

To calculate the capacity of a refrigeration plant, use the total heat removed and account for the efficiency of the system.

Answer 1

Correct Answer: 2.2

The total heat removed from the potato is: \[ Q_{\text{total}} = 5000 \cdot 3.65 \cdot (28 - 2) = 5000 \cdot 3.65 \cdot 26 = 475000 \, \text{kJ}. \] The net heat after considering the efficiency is: \[ Q_{\text{net}} = \frac{475000}{0.70} = 678571.43 \, \text{kJ}. \] Now, convert to tons of refrigeration: \[ 1 \, \text{ton of refrigeration} = 3.517 \times 10^3 \, \text{kJ/hr}. \] The capacity in ton of refrigeration is: \[ \text{Capacity} = \frac{678571.43}{3.517 \times 10^3} \approx 2.24 \, \text{tons}. \] Thus, the capacity of the plant is \( 2.24 \) tons of refrigeration.