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    A bucket has 101 water at 15°c. How much water at 50°c should be added to get a mixture of temperature 40°C? (Assume no heat loss in mixing.)
    Question

    A bucket has 101 water at 15°c. How much water at 50°c should be added to get a mixture of temperature 40°C? (Assume no heat loss in mixing.)

    A.

    15 L

    B.

    20 L

    C.

    25 L

    D.

    30 L

    Correct option is C

    Given:
    Initial water: 10 L at 15°C
    Added water: x L at 50°C
    Final mixture temperature: 40°C
    No heat loss (heat gained = heat lost).
    Concept Used:
    The principle of heat exchange states:

    Heat gained by cooler water = Heat lost by hotter water

    The heat gained or lost by a substance is given by:

    Q = m × c × ΔT

    where:

    m = mass (or volume in this case, since density of water is 1 kg/L)
    c = specific heat (same for water, so it cancels out)
    ΔT = change in temperature
    Thus, we set up the equation:

    (Mass of cold water) × (Temp increase) = (Mass of hot water) × (Temp decrease)

    Solution:

    Heat gained by cold water (10 L at 15°C):
    10 × (40 - 15) = 10 × 25 = 250

    Heat lost by hot water (x L at 50°C):
    x × (50 - 40) = x × 10 = 10x

    Setting heat gained = heat lost:

    250 = 10x

    Solving for x:

    x = 250 / 10 = 25 L

    Final Answer:
    (C) 25 L

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