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A milk chilling unit can remove heat from the milk at the rate of 42 MJ/h. The rate of heat leakage into the milk from the surrounding is 4.2 MJ/h. The specific heat of milk above the freezing point is 4.2 kJ/kg K and below the freezing point is 2.1 kJ/kg K. The freezing point of milk is -2 oc and the latent heat is 100 kJ/kg. The time required (in min) for cooling a batch of 100 kg milk from 47 oc to -5 oc is
    Correct answer is between '49,50'. Can you explain this answer?
    Verified Answer
    A milk chilling unit can remove heat from the milk at the rate of 42 M...
    Suppose the time required is t hrs then
    The amount of heat removed
    = heat removed to decrease the temperature from 47 oc to -2 oc + latent heat + heat removed to decrease the temperature from -2 oc to -5 oc
    (42000 – 4200)t = 100 ×[4.2 × (47+2) + 100 + 2.1× (−2+5)]
    t = 0.825 hr = 49.5 min
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    A milk chilling unit can remove heat from the milk at the rate of 42 M...
    Given data:
    - Rate of heat removal by milk chilling unit: 42 MJ/h
    - Rate of heat leakage into the milk from the surroundings: 4.2 MJ/h
    - Specific heat of milk above freezing point: 4.2 kJ/kg K
    - Specific heat of milk below freezing point: 2.1 kJ/kg K
    - Freezing point of milk: -2°C
    - Latent heat of milk: 100 kJ/kg
    - Batch size of milk: 100 kg
    - Initial temperature of milk: 47°C
    - Final temperature of milk: -5°C

    Approach:
    To determine the time required for cooling the milk batch from 47°C to -5°C, we need to consider the heat transfer processes involved. We can break down the process into the following stages:
    1. Cooling the milk from 47°C to the freezing point
    2. Freezing the milk at the freezing point
    3. Cooling the frozen milk from the freezing point to -5°C

    Stage 1: Cooling the milk from 47°C to the freezing point
    - Heat to be removed = Mass of milk x Specific heat (above freezing point) x Temperature difference
    - Heat to be removed = 100 kg x 4.2 kJ/kg K x (47°C - (-2°C))
    - Heat to be removed = 100 kg x 4.2 kJ/kg K x 49°C
    - Heat to be removed = 20580 kJ

    Stage 2: Freezing the milk at the freezing point
    - Heat to be removed = Mass of milk x Latent heat
    - Heat to be removed = 100 kg x 100 kJ/kg
    - Heat to be removed = 10000 kJ

    Stage 3: Cooling the frozen milk from the freezing point to -5°C
    - Heat to be removed = Mass of milk x Specific heat (below freezing point) x Temperature difference
    - Heat to be removed = 100 kg x 2.1 kJ/kg K x ((-5°C) - (-2°C))
    - Heat to be removed = 100 kg x 2.1 kJ/kg K x 3°C
    - Heat to be removed = 630 kJ

    Total heat to be removed:
    Total heat to be removed = Heat from Stage 1 + Heat from Stage 2 + Heat from Stage 3
    Total heat to be removed = 20580 kJ + 10000 kJ + 630 kJ
    Total heat to be removed = 31210 kJ

    Time required:
    Time required = Total heat to be removed / Rate of heat removal
    Time required = 31210 kJ / (42 MJ/h - 4.2 MJ/h)
    Time required = 31210 kJ / 37.8 MJ/h
    Time required = 0.824 h

    Converting hours to minutes:
    Time required = 0.824 h x 60 min/h
    Time required = 49.44 min

    Therefore, the time required for cooling a batch of 100 kg milk from 47°C to -5°C is approximately 49 minutes.
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    A milk chilling unit can remove heat from the milk at the rate of 42 MJ/h. The rate of heat leakage into the milk from the surrounding is 4.2 MJ/h. The specific heat of milk above the freezing point is 4.2 kJ/kg K and below the freezing point is 2.1 kJ/kg K. The freezing point of milk is -2ocand the latent heat is 100 kJ/kg. The time required (in min) for cooling a batch of 100 kg milk from 47oc to -5ocisCorrect answer is between '49,50'. Can you explain this answer?
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    A milk chilling unit can remove heat from the milk at the rate of 42 MJ/h. The rate of heat leakage into the milk from the surrounding is 4.2 MJ/h. The specific heat of milk above the freezing point is 4.2 kJ/kg K and below the freezing point is 2.1 kJ/kg K. The freezing point of milk is -2ocand the latent heat is 100 kJ/kg. The time required (in min) for cooling a batch of 100 kg milk from 47oc to -5ocisCorrect answer is between '49,50'. Can you explain this answer? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about A milk chilling unit can remove heat from the milk at the rate of 42 MJ/h. The rate of heat leakage into the milk from the surrounding is 4.2 MJ/h. The specific heat of milk above the freezing point is 4.2 kJ/kg K and below the freezing point is 2.1 kJ/kg K. The freezing point of milk is -2ocand the latent heat is 100 kJ/kg. The time required (in min) for cooling a batch of 100 kg milk from 47oc to -5ocisCorrect answer is between '49,50'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A milk chilling unit can remove heat from the milk at the rate of 42 MJ/h. The rate of heat leakage into the milk from the surrounding is 4.2 MJ/h. The specific heat of milk above the freezing point is 4.2 kJ/kg K and below the freezing point is 2.1 kJ/kg K. The freezing point of milk is -2ocand the latent heat is 100 kJ/kg. The time required (in min) for cooling a batch of 100 kg milk from 47oc to -5ocisCorrect answer is between '49,50'. Can you explain this answer?.
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