The Question is from neet-physics thermodynamics , and it was created by the author NAGA SAI

Question

The temperature inside a refrigerator is ${t_2}{\,^{\rm{o}}}C$ and the room temperature is ${t_1}{\,^o}C$. The amount of heat delivered to the room for each joule of electrical energy consumed ideally will be

NAGA SAI · Accepted Answer

## CORE INFORMATION
✓ **Answer**: C
## CONCEPT FRAMEWORK
	**Primary Concept**: Coefficient of Performance (COP) of a refrigerator
	**Related Topics**: Thermodynamics, Heat Transfer, Temperature Scales
	**Prerequisites**: Understanding of COP, heat transfer principles, and temperature conversion between Celsius and Kelvin.
## SOLUTION PATHWAY
**Given Information**:
	* Refrigerator temperature: $t_2$ °C
	* Room temperature: $t_1$ °C
	*  We are looking for the heat delivered to the room per joule of electrical energy consumed. This is the COP.
**Method & Steps**:
	1. Convert Celsius temperatures to Kelvin: $T_1 = t_1 + 273$ and $T_2 = t_2 + 273$
		Reasoning: Thermodynamic calculations require absolute temperature (Kelvin).
		Formula/Rule: $T(K) = t(°C) + 273$
	2. Define the Coefficient of Performance (COP) for a refrigerator: $	ext{COP}_{refrigerator} = \frac{	ext{Heat extracted from cold reservoir}}{	ext{Work Input}} = \frac{Q_2}{W}$.
		Reasoning:  The COP represents the efficiency of the refrigerator.
	3. For an ideal refrigerator, the COP is also given by: $	ext{COP} = \frac{T_2}{T_1 - T_2}$. This relates the temperatures to the heat transfer.
		Reasoning: Ideal refrigerator concept and relationship with temperatures
	4. Since we want the heat delivered to the room ($Q_1$) per unit work, and $Q_1 = Q_2 + W$ (Conservation of Energy), we can say that $\frac{Q_1}{W}$ is the required value.
		Reasoning: Conservation of energy and heat delivered to the hot reservoir.
	5. We have $\frac{Q_2}{W} = \frac{T_2}{T_1 - T_2}$, and $Q_1=Q_2+W$. So, $Q_1 = \frac{T_2}{T_1 - T_2}W + W = W\frac{T_1}{T_1-T_2}$. Therefore, $\frac{Q_1}{W} = \frac{T_1}{T_1-T_2}$
		Reasoning:  Relating heat delivered to the room to the COP and the work done. 
	6. Substituting the Kelvin values: $\frac{Q_1}{W} = \frac{t_1 + 273}{(t_1 + 273) - (t_2 + 273)} = \frac{t_1 + 273}{t_1 - t_2}$
		Result:  The amount of heat delivered to the room per joule of electrical energy is $\frac{t_1 + 273}{t_1 - t_2}$.
		Verification: The derived expression matches option C.
 ## OPTION ANALYSIS
[A] :x: Incorrect. This option does not have the correct temperature difference in the denominator.
[B] :x: Incorrect. This option does not include the temperature conversion to Kelvin and the correct ratio between temperatures.
[C] ✓: Correct. This option has the correct formula for the heat delivered to the hot reservoir per joule of energy consumed.
[D] :x: Incorrect. This option has the incorrect temperatures in the numerator and denominator.
 ## LEARNING AIDS
	**Common Mistakes**:
		1. [Error 1]: Forgetting to convert Celsius to Kelvin. This will lead to an incorrect result. Always use absolute temperature in thermodynamic calculations.
		2. [Error 2]: Confusing the COP for a refrigerator with that of a heat engine. The formula is different, and the objective of the device is different.
	**Quick Tips**:
		1. Remember that the COP is the ratio of what you want (heat extracted or delivered) over what you pay (work input).
		2. For ideal refrigerators, the COP is directly related to absolute temperatures (Kelvin).

NAGA SAI · Answer

$\frac{{{t_1} + {t_2}}}{{{t_1} + 273}}$

NAGA SAI · Answer

$\frac{{{t_1}}}{{{t_1} - {t_2}}}$

NAGA SAI · Answer

The correct Option is "$\frac{{{t_1} + 273}}{{{t_1} - {t_2}}}$"

NAGA SAI · Answer

$\frac{{{t_2} + 273}}{{{t_1} + {t_2}}}$

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