The temperature gradient in a rod of 0.5 m l | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The temperature gradient is defined as the change in temperature per unit length,given by the formula: $	ext{Temperature Gradient} = \frac{	heta

Vedclass · Accepted Answer

$-10$

Vedclass · Answer

(B) The temperature gradient is defined as the change in temperature per unit length,given by the formula: $	ext{Temperature Gradient} = \frac{	heta_1 - 	heta_2}{l}$.Here,$	heta_1 = 30 \ ^oC$ is the temperature of the hotter end,$l = 0.5 \ m$ is the length of the rod,and the temperature gradient is $80 \ ^oC/m$.Substituting the values into the formula:$80 = \frac{30 - 	heta_2}{0.5}$.Multiplying both sides by $0.5$:$80 	imes 0.5 = 30 - 	heta_2$.$40 = 30 - 	heta_2$.Rearranging to solve for $	heta_2$:$	heta_2 = 30 - 40 = -10 \ ^oC$.Therefore,the temperature of the cooler end is $-10 \ ^oC$.

The temperature gradient in a rod of $0.5 \ m$ long is $80 \ ^oC/m$. If the temperature of the hotter end of the rod is $30 \ ^oC$,then the temperature of the cooler end is ...... $^oC$.

Similar Questions

In the steady state,the temperature of a body ...

$A$ thermocouple is an arrangement of two different metals to :-

What is heat transfer? Write its modes. In which mode,medium is not required?

Define heat transfer in solids.

Vedclass Test Series

Exam Paper Generator

Online Exam Module