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A container with 1 kg of water in it is kept in sunlight, which causes the water to get warmer than the surroundings. The average energy per unit time per unit area received due to the sunlight is 700Wm−2 and it is absorbed by the water over an effective area of 0.05 m2. Assuming that the heat loss from the water to the surroundings is governed by Newton’s law of cooling, the difference (in °C) in the temperature of water and the surroundings after a long time will be _____________. (Ignore effect of the container, and take constant for Newton’s law of cooling = 0.001 s−1, Heat capacity of water = 4200 J kg−1 K−1) 
    Correct answer is '8.33'. Can you explain this answer?
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    A container with 1 kg of water in it is kept in sunlight, which causes...




    Now from equ. (i) 



    ΔT = 8.33
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    A container with 1 kg of water in it is kept in sunlight, which causes...
    -2. As a result, the container and the water inside it absorb some of this energy, which causes the temperature of the water to increase.

    The amount of energy absorbed by the water can be calculated using the formula:

    E = mcΔT

    Where:
    E = energy absorbed by the water (in Joules)
    m = mass of water (in kg)
    c = specific heat capacity of water (4.18 J/g°C or 4180 J/kg°C)
    ΔT = change in temperature of water (in °C)

    In this case, m = 1 kg, c = 4180 J/kg°C, and let's assume that the temperature of the water increases by 10°C:

    ΔT = 10°C

    Plugging these values into the formula, we get:

    E = 1 kg x 4180 J/kg°C x 10°C
    E = 41,800 J

    The amount of energy absorbed by the water is therefore 41,800 Joules.
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    The atmospheric lapse rateFor small volumes of gas, according to kinetic theory of gases, all parts of the gas are at the same temperature. But for huge volumes of gas like atmosphere, assumption of a uniform temperature throughout the gas is not valid. Different parts of the atmosphere are at different temperatures. Apart from the surface of the earth, variations also occur in temperature at different heights in the atmosphere.The decrease in temperature with height called the atmospheric lapse rate is similar at various locations across the surface of the Earth. By analyzing the data collected at various locations, it is found that average global lapse rate is – 6.7 °C/Km.The linear decrease with temperature only occurs in the lower part of the atmosphere called the troposphere. This is the part of the atmosphere in which weather occurs and our planes fly. Above the troposphere is the stratosphere, with an imaginary boundary separating the two layers. In the stratosphere, temperature tends to be relatively constant.Absorption of sunlight at the Earth’s surface warms the troposphere from below, so vertical convection currents are continually mixing in the air. As a parcel of air rises, its pressure drops and it expands. The parcel does work on its surrounding, so that its internal energy and therefore, its temperature drops. Assume that the vertical mixing is so rapid as to be adiabatic and the quantityTP(1 – λ)/λ has a uniform value through the layers of troposphere.(M is molecular mass of the air, R is universal gas constant, g is gravitational acc., P and T are pressure and temperature respectively at the point under consideration and y is height.)Q. The value of theoretical lapse rate on the earth is (use g = 9.8 m/s2 ; R = 8.3 J/mol-k and M = 29 g/mol)

    The atmospheric lapse rateFor small volumes of gas, according to kinetic theory of gases, all parts of the gas are at the same temperature. But for huge volumes of gas like atmosphere, assumption of a uniform temperature throughout the gas is not valid. Different parts of the atmosphere are at different temperatures. Apart from the surface of the earth, variations also occur in temperature at different heights in the atmosphere.The decrease in temperature with height called the atmospheric lapse rate is similar at various locations across the surface of the Earth. By analyzing the data collected at various locations, it is found that average global lapse rate is – 6.7 °C/Km.The linear decrease with temperature only occurs in the lower part of the atmosphere called the troposphere. This is the part of the atmosphere in which weather occurs and our planes fly. Above the troposphere is the stratosphere, with an imaginary boundary separating the two layers. In the stratosphere, temperature tends to be relatively constant.Absorption of sunlight at the Earth’s surface warms the troposphere from below, so vertical convection currents are continually mixing in the air. As a parcel of air rises, its pressure drops and it expands. The parcel does work on its surrounding, so that its internal energy and therefore, its temperature drops. Assume that the vertical mixing is so rapid as to be adiabatic and the quantityTP(1 – λ)/λ has a uniform value through the layers of troposphere.(M is molecular mass of the air, R is universal gas constant, g is gravitational acc., P and T are pressure and temperature respectively at the point under consideration and y is height.)Q. Mechanical equilibrium of the atmosphere requires that the pressure decreases with altitude according to . Assuming free fall acceleration to be uniform, then lapse rate is given by

    The atmospheric lapse rateFor small volumes of gas, according to kinetic theory of gases, all parts of the gas are at the same temperature. But for huge volumes of gas like atmosphere, assumption of a uniform temperature throughout the gas is not valid. Different parts of the atmosphere are at different temperatures. Apart from the surface of the earth, variations also occur in temperature at different heights in the atmosphere.The decrease in temperature with height called the atmospheric lapse rate is similar at various locations across the surface of the Earth. By analyzing the data collected at various locations, it is found that average global lapse rate is – 6.7 °C/Km.The linear decrease with temperature only occurs in the lower part of the atmosphere called the troposphere. This is the part of the atmosphere in which weather occurs and our planes fly. Above the troposphere is the stratosphere, with an imaginary boundary separating the two layers. In the stratosphere, temperature tends to be relatively constant.Absorption of sunlight at the Earth’s surface warms the troposphere from below, so vertical convection currents are continually mixing in the air. As a parcel of air rises, its pressure drops and it expands. The parcel does work on its surrounding, so that its internal energy and therefore, its temperature drops. Assume that the vertical mixing is so rapid as to be adiabatic and the quantityTP(1 – λ)/λ has a uniform value through the layers of troposphere.(M is molecular mass of the air, R is universal gas constant, g is gravitational acc., P and T are pressure and temperature respectively at the point under consideration and y is height.)Q. If behaviour of the mixing of parcels of air is approximately assumed to be adiabatic then lapse rate can be expressed as

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    A container with 1 kg of water in it is kept in sunlight, which causes the water to get warmer than the surroundings. The average energy per unit time per unit area received due to the sunlight is 700Wm−2 and it is absorbed by the water over an effective area of 0.05 m2. Assuming that the heat loss from the water to the surroundings is governed by Newton’s law of cooling, the difference (in °C) in the temperature of water and the surroundings after a long time will be _____________. (Ignore effect of the container, and take constant for Newton’s law of cooling = 0.001 s−1, Heat capacity of water = 4200 J kg−1 K−1)Correct answer is '8.33'. Can you explain this answer?
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    A container with 1 kg of water in it is kept in sunlight, which causes the water to get warmer than the surroundings. The average energy per unit time per unit area received due to the sunlight is 700Wm−2 and it is absorbed by the water over an effective area of 0.05 m2. Assuming that the heat loss from the water to the surroundings is governed by Newton’s law of cooling, the difference (in °C) in the temperature of water and the surroundings after a long time will be _____________. (Ignore effect of the container, and take constant for Newton’s law of cooling = 0.001 s−1, Heat capacity of water = 4200 J kg−1 K−1)Correct answer is '8.33'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A container with 1 kg of water in it is kept in sunlight, which causes the water to get warmer than the surroundings. The average energy per unit time per unit area received due to the sunlight is 700Wm−2 and it is absorbed by the water over an effective area of 0.05 m2. Assuming that the heat loss from the water to the surroundings is governed by Newton’s law of cooling, the difference (in °C) in the temperature of water and the surroundings after a long time will be _____________. (Ignore effect of the container, and take constant for Newton’s law of cooling = 0.001 s−1, Heat capacity of water = 4200 J kg−1 K−1)Correct answer is '8.33'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A container with 1 kg of water in it is kept in sunlight, which causes the water to get warmer than the surroundings. The average energy per unit time per unit area received due to the sunlight is 700Wm−2 and it is absorbed by the water over an effective area of 0.05 m2. Assuming that the heat loss from the water to the surroundings is governed by Newton’s law of cooling, the difference (in °C) in the temperature of water and the surroundings after a long time will be _____________. (Ignore effect of the container, and take constant for Newton’s law of cooling = 0.001 s−1, Heat capacity of water = 4200 J kg−1 K−1)Correct answer is '8.33'. Can you explain this answer?.
    Solutions for A container with 1 kg of water in it is kept in sunlight, which causes the water to get warmer than the surroundings. The average energy per unit time per unit area received due to the sunlight is 700Wm−2 and it is absorbed by the water over an effective area of 0.05 m2. Assuming that the heat loss from the water to the surroundings is governed by Newton’s law of cooling, the difference (in °C) in the temperature of water and the surroundings after a long time will be _____________. (Ignore effect of the container, and take constant for Newton’s law of cooling = 0.001 s−1, Heat capacity of water = 4200 J kg−1 K−1)Correct answer is '8.33'. Can you explain this answer? in English & in Hindi are available as part of our courses for JEE. Download more important topics, notes, lectures and mock test series for JEE Exam by signing up for free.
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