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 700Wmandminus;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 Newtonandrsquo;s law of cooling, the difference (in anddeg;C) in the temperature of water and the surroundings after a long time will be _____________. (Ignore effect of the container, and take constant for Newtonandrsquo;s law of cooling = 0.001 sandminus;1, Heat capacity of water = 4200 J kgandminus;1 Kandminus;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?

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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.

Answered by Infinity Academy · View profile

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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?.

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