Question

A milk chilling unit can remove heat from the milk at the rate of 41.87 MJ/h. Heatleaks into the milk from the surroundings at an average rate of 4.187 MJ/h. Find thetime required for cooling a batch of 500 kg of milk from 45°C to 5°C. Take the cp ofmilk to be 4.187 kj/kg K.​

Answer 1

Explanation:

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Answer 2

$t=133.33\ minutes$ is the time required for cooling the milk.

Explanation:

Given:

• rate of cooling the milk, $H_{out}=4.187\times 10^7\ J.hr^{-1}$

• rate of heat leakage, $H_{in}=4.187\times 10^6\ J.hr^{-1}$

• mass of milk to be cooled, $m=500\ kg$

• initial temperature of the milk, $T_i=45^{\circ}C$

• initial temperature of the milk, $T_f=5^{\circ}C$

• specific heat capacity of the milk,  $c=4187\ J.kg^{-1}.K{-1}$

Now the net heat removal rate compensating the heat leakage:

$H=37.683\times 10^6\ J.hr^{-1}$

Total heat of the milk to be removed:

$Q=m.c.\Delta T$

$Q=500\times 4187\times (45-5)$

$Q=8.374\times 10^7\ J$

Now the time required for cooling:

$t=\frac{Q}{H}$

$t=\frac{8.374\times 10^7}{3.7683\times 10^7}$

$t=133.33\ minutes$

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TOPIC:specific heat capacity, Power

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