A cold storage takes 5 hours to bring down the temperature of 100 metric tons of potato from $35^{\circ} \mathrm{C}$ to $8^{\circ} \mathrm{C}$. The specific heat capacity of potato is $3.1 \mathrm{~kJ} \mathrm{~kg}^{-1}{ }^{\circ} \mathrm{C}^{-1} .$ The coefficient of performance (COP) and the latent heat of vapourisation of the refrigerant $(\mathrm{R}-22)$ at an evaporation temperature of $-10^{\circ} \mathrm{C}$ are 3.66 and $230 \mathrm{~kJ} \mathrm{~kg}^{-1},$ respectively. Neglecting respiration heat load of potato, and assuming no power loss, the values of refrigerant flow rate and the power input to the compressor are
(A) $121.3 \mathrm{~kg} \min ^{-1}$ and $127.1 \mathrm{~kW}$
(B) $124.7 \mathrm{~kg} \mathrm{~min}^{-1}$ and $121.3 \mathrm{~kW}$
(C) $127.1 \mathrm{~kg} \min ^{-1}$ and $121.3 \mathrm{~kW}$
(D) $124.7 \mathrm{~kg} \min ^{-1}$ and $127.1 \mathrm{~kW}$
A cold storage takes 5 hours to bring down the temperature of 100 metric tons of potato from $35^{\circ} \mathrm{C}$ to $8^{\circ} \mathrm{C}$. The specific heat capacity of potato is $3.1 \mathrm{~kJ} \mathrm{~kg}^{-1}{ }^{\circ} \mathrm{C}^{-1} .$ The coefficient of performance (COP) and the latent heat of vapourisation of the refrigerant $(\mathrm{R}-22)$ at an evaporation temperature of $-10^{\circ} \mathrm{C}$ are 3.66 and $230 \mathrm{~kJ} \mathrm{~kg}^{-1},$ respectively. Neglecting respiration heat load of potato, and assuming no power loss, the values of refrigerant flow rate and the power input to the compressor are
(A) $121.3 \mathrm{~kg} \min ^{-1}$ and $127.1 \mathrm{~kW}$
(B) $124.7 \mathrm{~kg} \mathrm{~min}^{-1}$ and $121.3 \mathrm{~kW}$
(C) $127.1 \mathrm{~kg} \min ^{-1}$ and $121.3 \mathrm{~kW}$
(D) $124.7 \mathrm{~kg} \min ^{-1}$ and $127.1 \mathrm{~kW}$
(A) $121.3 \mathrm{~kg} \min ^{-1}$ and $127.1 \mathrm{~kW}$