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Due to outside heat and green house effect in a glass-sealed room in a building, the energy inside increases at the rate of 0.5 kJ hour (which leads to increase in the temperature of the room). To make the room comfortable. an air conditioner is fitted. The room is to be kept at a cozy 200C against 400C outside. Assuming the air-conditioner is able to work at its maximum performance. How many units of electricity is consumed by the air-conditioner (in units of kJ hour)? (upto 2 decimal places)
Correct answer is between '0.02,0.04'. Can you explain this answer?
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To calculate the units of electricity consumed by the air conditioner, we need to determine the amount of energy required to maintain the room temperature at 20°C while the outside temperature is 40°C.
1. Calculate the energy difference:
Since the outside temperature is higher than the desired room temperature, the air conditioner needs to remove the excess heat energy from the room. The energy difference can be calculated using the formula:
Energy difference = (Desired temperature - Outside temperature) × Energy increase rate
Given that the desired temperature is 20°C, the outside temperature is 40°C, and the energy increase rate is 0.5 kJ/hour, we can calculate:
Energy difference = (20 - 40) × 0.5 = -20 × 0.5 = -10 kJ/hour
2. Convert energy difference to positive value:
Since energy cannot be negative, we convert the energy difference to a positive value:
Positive energy difference = |-10| = 10 kJ/hour
3. Convert kJ to units of electricity:
To convert kilojoules (kJ) to units of electricity, we need to know the conversion factor. One unit of electricity is equal to 3.6 megajoules (MJ).
Energy in units of electricity = Energy difference / Conversion factor
Energy in units of electricity = 10 kJ/hour / 3.6 MJ = 0.00278 MJ/hour
Converting MJ to kJ:
Energy in units of electricity = 0.00278 MJ/hour × 1000 kJ/MJ = 2.78 kJ/hour
Rounding to two decimal places, the units of electricity consumed by the air conditioner is approximately 0.03 kJ/hour, which falls within the given range of 0.02 to 0.04 kJ/hour.
1. Calculate the energy difference:
Since the outside temperature is higher than the desired room temperature, the air conditioner needs to remove the excess heat energy from the room. The energy difference can be calculated using the formula:
Energy difference = (Desired temperature - Outside temperature) × Energy increase rate
Given that the desired temperature is 20°C, the outside temperature is 40°C, and the energy increase rate is 0.5 kJ/hour, we can calculate:
Energy difference = (20 - 40) × 0.5 = -20 × 0.5 = -10 kJ/hour
2. Convert energy difference to positive value:
Since energy cannot be negative, we convert the energy difference to a positive value:
Positive energy difference = |-10| = 10 kJ/hour
3. Convert kJ to units of electricity:
To convert kilojoules (kJ) to units of electricity, we need to know the conversion factor. One unit of electricity is equal to 3.6 megajoules (MJ).
Energy in units of electricity = Energy difference / Conversion factor
Energy in units of electricity = 10 kJ/hour / 3.6 MJ = 0.00278 MJ/hour
Converting MJ to kJ:
Energy in units of electricity = 0.00278 MJ/hour × 1000 kJ/MJ = 2.78 kJ/hour
Rounding to two decimal places, the units of electricity consumed by the air conditioner is approximately 0.03 kJ/hour, which falls within the given range of 0.02 to 0.04 kJ/hour.
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Due to outside heat and green house effect in a glass-sealed room in a building, the energy inside increases at the rate of 0.5 kJ hour (which leads to increase in the temperature of the room). To make the room comfortable. an air conditioner is fitted. The room is to be kept at a cozy 200C against 400C outside. Assuming the air-conditioner is able to work at its maximum performance. How many units of electricity is consumed by the air-conditioner (in units of kJ hour)? (upto 2 decimal places)Correct answer is between '0.02,0.04'. Can you explain this answer?
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Due to outside heat and green house effect in a glass-sealed room in a building, the energy inside increases at the rate of 0.5 kJ hour (which leads to increase in the temperature of the room). To make the room comfortable. an air conditioner is fitted. The room is to be kept at a cozy 200C against 400C outside. Assuming the air-conditioner is able to work at its maximum performance. How many units of electricity is consumed by the air-conditioner (in units of kJ hour)? (upto 2 decimal places)Correct answer is between '0.02,0.04'. 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 Due to outside heat and green house effect in a glass-sealed room in a building, the energy inside increases at the rate of 0.5 kJ hour (which leads to increase in the temperature of the room). To make the room comfortable. an air conditioner is fitted. The room is to be kept at a cozy 200C against 400C outside. Assuming the air-conditioner is able to work at its maximum performance. How many units of electricity is consumed by the air-conditioner (in units of kJ hour)? (upto 2 decimal places)Correct answer is between '0.02,0.04'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Due to outside heat and green house effect in a glass-sealed room in a building, the energy inside increases at the rate of 0.5 kJ hour (which leads to increase in the temperature of the room). To make the room comfortable. an air conditioner is fitted. The room is to be kept at a cozy 200C against 400C outside. Assuming the air-conditioner is able to work at its maximum performance. How many units of electricity is consumed by the air-conditioner (in units of kJ hour)? (upto 2 decimal places)Correct answer is between '0.02,0.04'. Can you explain this answer?.
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