Absolute Zero on celsius scale

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SUMMARY

The discussion centers on determining the location of absolute zero on the Celsius scale using the Carnot cycle. Arya presented a problem involving a heat input of 20 KJ at 100°C and a heat rejection of 14.6 KJ at 0°C. The solution derived is -270.16°C, achieved by applying the Carnot cycle efficiency formula and understanding the relationship between heat input, heat rejection, and the temperatures of the reservoirs. The key takeaway is the necessity to treat absolute zero as a variable in thermodynamic calculations.

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arya_jsr
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Hi ,
Please help me to solve following problem.

If 20 KJ are added to a carnot cycle at a temperature 100 degree Celsius and 14.6 KJ are rejected at 0 degree Celsius,determine the location of absolute zero on celsius scale?

Ans -270.16 C

Please tell me how to solve it

Thanks in advance.
Regards
Arya
 
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Hi Arya, welcome to PF. What do you know about the Carnot cycle relationships between extracted work, rejected heat, and the temperatures of the hot and cold reservoirs?
 
Mapes said:
Hi Arya, welcome to PF. What do you know about the Carnot cycle relationships between extracted work, rejected heat, and the temperatures of the hot and cold reservoirs?

Hi Mapes,
In carnot cycle suppose Q1 is given as heat input and T1 is temperature of source,and
Q2 is rejected heat at temperature T2(Temperature of sink),then (Q1/T1)=(Q2/T2) and if W is the work done by reversible cyclic heat engine then Q1=Q2+W.
Now please tell me how to solve this problem
regards
Arya
 
The Efficiency is defined as:

E=\frac{Q_h-Q_c}{Q_h}

Qh is the heat extracted and Qc is the head expelled.

Then, we can derive using basic thermodynamic laws that the efficiency of the Carnot Cycle is given by:

E=1-\frac{T_c}{T_h}

Where Th is the temperature of the hot reservoir and Tc is the temperature of the cold reservoir. Can you figure it out from here?
 
arya_jsr said:
Hi Mapes,
In carnot cycle suppose Q1 is given as heat input and T1 is temperature of source,and
Q2 is rejected heat at temperature T2(Temperature of sink),then (Q1/T1)=(Q2/T2) and if W is the work done by reversible cyclic heat engine then Q1=Q2+W.

Looks good. Now T_1 and T_2 are measured from the absolute zero of the system, which is a variable for this problem. So you can't just use 100°C and 0°C; rather, you need to reference these to the absolute zero variable. Then solve the equation for this variable.
 
Thanks My Friend Mapes,
I got it ,it solved my problem.Actually i was not taking absolute zero as a variable .Now i have solved it.Thanks
regards
Arya
 

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