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Certain quantity of water cools from 70°C to 60°C in the first 5 minutes and to 54° C in the next 5 minutes. The temperature of the surroundings is
- a)42°C
- b)10°C
- c)45°C
- d)20°C
Correct answer is option 'C'. Can you explain this answer?
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Certain quantity of water cools from 70°C to 60°C in the first...
°F to 50°F in a room with an ambient temperature of 80°F. The heat exchange between the water and the room can be calculated using the formula:
Q = mcΔT
Where:
Q = heat exchange
m = mass of water
c = specific heat capacity of water
ΔT = change in temperature
First, we need to determine the mass of water. Assuming a density of 1 g/cm³ for water, we can calculate the volume of water using the formula:
V = (mass of water) / (density of water)
Let's assume we have 1 liter of water, which is equivalent to 1000 cm³. Therefore, the mass of water would be:
m = V * density of water
m = 1000 cm³ * 1 g/cm³
m = 1000 g
Next, we calculate the heat exchange using the formula:
Q = mcΔT
Q = (1000 g) * (4.184 J/g°C) * (70°F - 50°F)
Q = 1000 * 4.184 * 20
Q = 83680 J
Therefore, the heat exchange between the water and the room is 83680 Joules.
Q = mcΔT
Where:
Q = heat exchange
m = mass of water
c = specific heat capacity of water
ΔT = change in temperature
First, we need to determine the mass of water. Assuming a density of 1 g/cm³ for water, we can calculate the volume of water using the formula:
V = (mass of water) / (density of water)
Let's assume we have 1 liter of water, which is equivalent to 1000 cm³. Therefore, the mass of water would be:
m = V * density of water
m = 1000 cm³ * 1 g/cm³
m = 1000 g
Next, we calculate the heat exchange using the formula:
Q = mcΔT
Q = (1000 g) * (4.184 J/g°C) * (70°F - 50°F)
Q = 1000 * 4.184 * 20
Q = 83680 J
Therefore, the heat exchange between the water and the room is 83680 Joules.
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Certain quantity of water cools from 70°C to 60°C in the first 5 minutes and to 54° C in thenext 5 minutes. The temperature of the surroundings isa)42°Cb)10°Cc)45°Cd)20°CCorrect answer is option 'C'. Can you explain this answer?
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Certain quantity of water cools from 70°C to 60°C in the first 5 minutes and to 54° C in thenext 5 minutes. The temperature of the surroundings isa)42°Cb)10°Cc)45°Cd)20°CCorrect answer is option 'C'. Can you explain this answer? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about Certain quantity of water cools from 70°C to 60°C in the first 5 minutes and to 54° C in thenext 5 minutes. The temperature of the surroundings isa)42°Cb)10°Cc)45°Cd)20°CCorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Certain quantity of water cools from 70°C to 60°C in the first 5 minutes and to 54° C in thenext 5 minutes. The temperature of the surroundings isa)42°Cb)10°Cc)45°Cd)20°CCorrect answer is option 'C'. Can you explain this answer?.
Certain quantity of water cools from 70°C to 60°C in the first 5 minutes and to 54° C in thenext 5 minutes. The temperature of the surroundings isa)42°Cb)10°Cc)45°Cd)20°CCorrect answer is option 'C'. Can you explain this answer? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about Certain quantity of water cools from 70°C to 60°C in the first 5 minutes and to 54° C in thenext 5 minutes. The temperature of the surroundings isa)42°Cb)10°Cc)45°Cd)20°CCorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Certain quantity of water cools from 70°C to 60°C in the first 5 minutes and to 54° C in thenext 5 minutes. The temperature of the surroundings isa)42°Cb)10°Cc)45°Cd)20°CCorrect answer is option 'C'. Can you explain this answer?.
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