The Question is from neet-physics thermal-properties-of-matter newtons-law-of-cooling, and it was created by the author Mr. ANIL KUMAR.G

Question

A spherical body of radius $r$ at temperature $T_1$ is placed in an environment at temperature $T_2$ ($T_2 < T_1$).  The body has emissivity $e$ and specific heat capacity $c$.  If the density of the body is $\rho$ and Stefan's constant is $\sigma$, which expression most accurately represents the initial rate of temperature change ($\frac{dT}{dt}$) of the body, considering both radiation and convection, assuming a constant convection coefficient $h$?

Mr. ANIL KUMAR.G · Accepted Answer

## CORE INFORMATION
✓ **Answer**: A
## CONCEPT FRAMEWORK
	**Primary Concept**: Heat Transfer (Radiation and Convection)
	**Related Topics**: Stefan-Boltzmann Law, Newton's Law of Cooling
	**Prerequisites**: Calculus, Thermodynamics
## SOLUTION PATHWAY
**Given Information**:
	* Radius: $r$
	* Initial Temperature: $T_1$
	* Ambient Temperature: $T_2$ ($T_2 < T_1$)
	* Emissivity: $e$
	* Specific Heat Capacity: $c$
	* Density: $ho$
	* Stefan's Constant: $\sigma$
	* Convection Coefficient: $h$
**Method & Steps**:
	1. **Heat Loss due to Radiation**: 
		Reasoning: The body radiates heat according to the Stefan-Boltzmann Law.
		Formula/Rule: $P_{rad} = e\sigma A(T_1^4 - T_2^4)$, where $A = 4\pi r^2$ is the surface area.
	2. **Heat Loss due to Convection**:
		Working: The body loses heat through convection according to Newton's Law of Cooling.
		Formula/Rule: $P_{conv} = hA(T_1 - T_2)$.
	3. **Total Heat Loss and Rate of Temperature Change**:
		Reasoning: The total heat loss is the sum of radiation and convection losses. This heat loss results in a decrease in the body's internal energy, which is related to the temperature change.
		Formula/Rule: $P_{total} = P_{rad} + P_{conv} = \frac{dQ}{dt} = mc\frac{dT}{dt}$, where $m = ho V = ho (\frac{4}{3}\pi r^3)$ is the mass.
		Working: $\frac{dT}{dt} = \frac{P_{total}}{mc} = \frac{e\sigma 4\pi r^2(T_1^4 - T_2^4) + h4\pi r^2(T_1 - T_2)}{ho (\frac{4}{3}\pi r^3)c} = \frac{3[e\sigma(T_1^4 - T_2^4) + h(T_1 - T_2)]}{rho c}$.
		Key Point: Since $T_1 > T_2$, the body is losing heat, so $\frac{dT}{dt}$ is negative.
		Result: $\frac{dT}{dt} = -\frac{3[e\sigma(T_1^4 - T_2^4) + h(T_1 - T_2)]}{rho c}$
		Verification: The expression has the correct units and the negative sign indicates cooling.
 ## OPTION ANALYSIS
[A] ✓: $-3[e\sigma(T_1^4 - T_2^4) + h(T_1 - T_2)] / (rho c)$ - Correct. This option includes both radiation and convection losses and the negative sign accounts for the cooling effect.
[B] :x: $3[e\sigma(T_1^4 - T_2^4) + h(T_1 - T_2)] / (rho c)$ - Incorrect. Missing the negative sign.
[C] :x: $-3[e\sigma(T_1^4 - T_2^4) - h(T_1 - T_2)] / (rho c)$ - Incorrect. The convection term should be added, not subtracted.
[D] :x: $3[e\sigma(T_1 - T_2) + h(T_1^4 - T_2^4)] / (rho c)$ - Incorrect. The temperature terms are incorrectly applied to radiation and convection.
 ## LEARNING AIDS
	**Common Mistakes**:
		1. Incorrect Sign: Forgetting the negative sign when the body is cooling.
			How to avoid: Consider the direction of heat flow.
		2. Confusing $T^4$ and $T$: Applying $T^4$ to convection and $T$ to radiation.
			How to avoid: Remember Stefan-Boltzmann Law for radiation and Newton's Law of Cooling for convection.
	**Quick Tips**:
		1. Dimensional Analysis: Check the units of the final expression.
		2. Sign Convention: Be mindful of the sign of $\frac{dT}{dt}$ based on whether the body is heating or cooling.

Mr. ANIL KUMAR.G · Answer

The correct Option is "-3[eσ(T_1^4 - T_2^4) + h(T_1 - T_2)] / (rρc)"

Mr. ANIL KUMAR.G · Answer

3[eσ(T_1^4 - T_2^4) + h(T_1 - T_2)] / (rρc)

Mr. ANIL KUMAR.G · Answer

-3[eσ(T_1^4 - T_2^4) - h(T_1 - T_2)] / (rρc)

Mr. ANIL KUMAR.G · Answer

3[eσ(T_1 - T_2) + h(T_1^4 - T_2^4)] / (rρc)

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