Question 782: Thermodynamics & KTG

Question

The temperature gradient in a rod of length $$75 \mathrm{~cm}$$ is $$40^{\circ} \mathrm{C} / \mathrm{m}$$. If the temperature of cooler end of the rod is $$10^{\circ} \mathrm{C}$$, then the temperature of hotter end is

Accepted Answer

We know
$$\begin{aligned}& \mathrm{T}_{\mathrm{g}}=\frac{\mathrm{T}_1-\mathrm{T}_2}{\mathrm{x}} \
& \Rightarrow \frac{\mathrm{T}_1-10}{0.75}=40
\end{aligned}$$
$$	herefore \quad$$ Temperature of the hotter end is:
$$\begin{aligned}
& \mathrm{T}_1=(40 	imes 0.75)+10 \
& \mathrm{~T}_1=40^{\circ} \mathrm{C}
\end{aligned}$$

Answer

$$50^{\circ} \mathrm{C}$$

Answer

$$40^{\circ} \mathrm{C}$$

Answer

$$35^{\circ} \mathrm{C}$$

Answer

$$25^{\circ} \mathrm{C}$$

Thermodynamics & KTG

The temperature gradient in a rod of length $75 \mathrm{~cm} is 40^{\circ} \mathrm{C} / \mathrm{m}. If the temperature of cooler end of the rod is 10^{\circ} \mathrm{C}$, then the temperature of hotter end is

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