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Ln a process at constant volume 24cal/gms of heat has been added to a system.Calculate the change in its temprature Ans-0.319T1K?
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Ln a process at constant volume 24cal/gms of heat has been added to a ...
To calculate the change in temperature when 24 cal/gms of heat is added to a system at constant volume, we can use the formula:
q = mcΔT
where q is the heat added, m is the mass of the system, c is the specific heat capacity of the system, and ΔT is the change in temperature.
In this case, the heat added is 24 cal/gms, and since we are given the heat per gram, we can assume the mass of the system is 1 gram. Therefore, m = 1g.
Next, we need to determine the specific heat capacity of the system. The specific heat capacity is the amount of heat required to raise the temperature of 1 gram of a substance by 1 degree Celsius.
Since the specific heat capacity is not given, we can assume it to be the specific heat capacity of water, which is approximately 1 cal/g°C.
Now we can substitute the values into the formula and solve for ΔT:
24 cal/gms = 1g x 1 cal/g°C x ΔT
Simplifying the equation, we have:
ΔT = 24 cal/gms / (1g x 1 cal/g°C)
ΔT = 24°C
Therefore, the change in temperature when 24 cal/gms of heat is added to the system at constant volume is 24°C.
Answer: ΔT = 24°C
q = mcΔT
where q is the heat added, m is the mass of the system, c is the specific heat capacity of the system, and ΔT is the change in temperature.
In this case, the heat added is 24 cal/gms, and since we are given the heat per gram, we can assume the mass of the system is 1 gram. Therefore, m = 1g.
Next, we need to determine the specific heat capacity of the system. The specific heat capacity is the amount of heat required to raise the temperature of 1 gram of a substance by 1 degree Celsius.
Since the specific heat capacity is not given, we can assume it to be the specific heat capacity of water, which is approximately 1 cal/g°C.
Now we can substitute the values into the formula and solve for ΔT:
24 cal/gms = 1g x 1 cal/g°C x ΔT
Simplifying the equation, we have:
ΔT = 24 cal/gms / (1g x 1 cal/g°C)
ΔT = 24°C
Therefore, the change in temperature when 24 cal/gms of heat is added to the system at constant volume is 24°C.
Answer: ΔT = 24°C
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Ln a process at constant volume 24cal/gms of heat has been added to a system.Calculate the change in its temprature Ans-0.319T1K?
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Ln a process at constant volume 24cal/gms of heat has been added to a system.Calculate the change in its temprature Ans-0.319T1K? for IIT JAM 2024 is part of IIT JAM preparation. The Question and answers have been prepared according to the IIT JAM exam syllabus. Information about Ln a process at constant volume 24cal/gms of heat has been added to a system.Calculate the change in its temprature Ans-0.319T1K? covers all topics & solutions for IIT JAM 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Ln a process at constant volume 24cal/gms of heat has been added to a system.Calculate the change in its temprature Ans-0.319T1K?.
Ln a process at constant volume 24cal/gms of heat has been added to a system.Calculate the change in its temprature Ans-0.319T1K? for IIT JAM 2024 is part of IIT JAM preparation. The Question and answers have been prepared according to the IIT JAM exam syllabus. Information about Ln a process at constant volume 24cal/gms of heat has been added to a system.Calculate the change in its temprature Ans-0.319T1K? covers all topics & solutions for IIT JAM 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Ln a process at constant volume 24cal/gms of heat has been added to a system.Calculate the change in its temprature Ans-0.319T1K?.
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