The Question is from neet-physics thermodynamics , and it was created by the author P.NARASIMHA

Question

A body cools from 50°C to 40°C in 5 minutes. The surrounding temperature is 20°C.  In the next 5 minutes, it will cool approximately to:

P.NARASIMHA · Accepted Answer

## CORE INFORMATION
✓ **Answer**: A
## CONCEPT FRAMEWORK
	**Primary Concept**: Newton's Law of Cooling
	**Related Topics**: Heat Transfer, Thermal Equilibrium
	**Prerequisites**: Basic Calculus, Temperature Measurement
## SOLUTION PATHWAY
**Given Information**:
	Initial temperature $T_1$ = 50°C
	Temperature after 5 minutes $T_2$ = 40°C
	Surrounding temperature $T_s$ = 20°C
	Time interval $Δt$ = 5 minutes
**Method & Steps**:
	1. **Apply Newton's Law of Cooling**
		Reasoning: The rate of cooling is proportional to the temperature difference between the body and its surroundings.
		Formula/Rule: $\frac{dT}{dt} = -k(T - T_s)$, where $k$ is a positive constant.
	2. **Calculate the average rate of cooling in the first 5 minutes**
		Working: $\frac{T_2 - T_1}{Δt} = \frac{40 - 50}{5} = -2$°C/min
		Key Point: This is an approximation since the rate of cooling changes continuously.
	3. **Approximate the temperature after the next 5 minutes**
		Working: Since the cooling rate is approximately constant over small intervals, we can use the average rate to estimate the temperature drop in the next 5 minutes: Temperature drop = $-2$°C/min * 5 min = -10°C. Therefore, the final temperature $T_3$ will be $T_2$ - 10°C = 40°C - 10°C = 30°C.
		Result:  However, the rate of cooling decreases as the temperature difference decreases.  Using Newton's Law of Cooling more precisely:
$\frac{dT}{dt} = -k(T - T_s)$ implies $\frac{dT}{T - T_s} = -k dt$. Integrating gives $ln|T - T_s| = -kt + C$.  Using the initial condition $T(0) = 50$, we get $ln|50 - 20| = C$, so $C = ln(30)$. Then $ln|T - 20| = -kt + ln(30)$.  When $t = 5$, $T = 40$, so $ln|40 - 20| = -5k + ln(30)$. Thus, $ln(20) = -5k + ln(30)$, and $k = \frac{ln(30) - ln(20)}{5} = \frac{ln(1.5)}{5}$.
Now, we want to find $T$ when $t = 10$. $ln|T - 20| = -10(\frac{ln(1.5)}{5}) + ln(30) = -2ln(1.5) + ln(30) = ln(\frac{30}{1.5^2}) = ln(\frac{30}{2.25}) = ln(13.33)$.  So, $T - 20 = 13.33$, and $T = 33.33$°C.
		Verification: The temperature difference between the body and the surroundings decreases over time, leading to a slower cooling rate.  The more precise calculation using Newton's Law of Cooling gives 33.33°C, which is closer to the correct answer.
## OPTION ANALYSIS
[A] ✓: 33.33°C - This is the correct answer obtained by applying Newton's Law of Cooling more precisely.
[B] :x: 35°C - This assumes a linear cooling rate, which is incorrect.
[C] :x: 36.67°C - This is even further from the correct answer and doesn't account for the decreasing cooling rate.
[D] :x: 32°C -  This underestimates the temperature and doesn't align with Newton's Law of Cooling.
## LEARNING AIDS
	**Common Mistakes**:
		1. Assuming a constant cooling rate: The rate of cooling decreases as the temperature difference between the body and the surroundings decreases.
			How to avoid: Apply Newton's Law of Cooling correctly.
		2. Incorrect application of logarithms: Be careful when manipulating logarithmic expressions.
			How to avoid: Review logarithm rules and practice solving related problems.
	**Quick Tips**:
		1. Remember that cooling is exponential, not linear.
		2. Newton's Law of Cooling is a differential equation; its solution involves logarithms.

P.NARASIMHA · Answer

The correct Option is "33.33°C"

P.NARASIMHA · Answer

35°C

P.NARASIMHA · Answer

36.67°C

P.NARASIMHA · Answer

32°C

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