The value of the temperature gradient is 80,^ | vedclass

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

Question

(A) The temperature gradient is defined as the rate of change of temperature with respect to distance, given by the formula: $\frac{T_{H} - T_{C}}{L}

Vedclass · Accepted Answer

$-10\,^{\circ}C$

Vedclass · Answer

(A) The temperature gradient is defined as the rate of change of temperature with respect to distance, given by the formula: $\frac{T_{H} - T_{C}}{L} = 80\,^{\circ}C/m$.Here, $T_{H} = 30\,^{\circ}C$ (temperature of the hot end), $L = 0.5\,m$ (length of the rod), and $T_{C}$ is the temperature of the cold end.Substituting the values into the equation:$\frac{30 - T_{C}}{0.5} = 80$$30 - T_{C} = 80 	imes 0.5$$30 - T_{C} = 40$$T_{C} = 30 - 40$$T_{C} = -10\,^{\circ}C$.Therefore, the temperature of the cold end is $-10\,^{\circ}C$.

The value of the temperature gradient is $80\,^{\circ}C/m$ on a rod of $0.5\,m$ length. If the temperature of the hot end is $30\,^{\circ}C$, what is the temperature of the cold end?

Similar Questions

Four rods of silver,copper,brass,and wood are of the same shape. They are heated together after wrapping a paper on each. The paper will burn first on:

What is thermal steady state?

Woollen clothes are used in winter season because woolen clothes

End $A$ of a copper rod is maintained in a steam chamber at $100\,^{\circ}C$ and the other end $B$ is maintained at $0\,^{\circ}C$. Assuming $x = 0$ at $A$,the $T-x$ graph in steady state will be:

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module