Question 916: Thermodynamics & KTG

Question

The rate of flow of heat through a copper rod with temperature difference $$28^{\circ} \mathrm{C}$$ is $$1400 \mathrm{~cal} \mathrm{~s}^{-1}$$. The thermal resistance of copper rod will be

Accepted Answer

The thermal resistance $R$ of a material is given by the ratio of the temperature difference ($\Delta T$) across the material to the rate of heat flow ($Q$) through the material. Mathematically, this can be expressed as:
$R = \frac{\Delta T}{Q}$
For this particular problem, we are given:
Temperature difference, $\Delta T = 28^{\circ} \mathrm{C}$
Rate of heat flow, $Q = 1400 \mathrm{~cal} \mathrm{~s}^{-1}$
Using the formula for thermal resistance, we can calculate $R$ as follows:
$R = \frac{28^{\circ} \mathrm{C}}{1400 \mathrm{~cal} \mathrm{~s}^{-1}}$
$R = 0.02 \frac{{ }^{\circ} \mathrm{C} \mathrm{s}}{\mathrm{cal}}$
Therefore, the thermal resistance of the copper rod is $0.02 \frac{{ }^{\circ} \mathrm{C} \mathrm{s}}{\mathrm{cal}}$. The correct answer is
Option B
.

Answer

$$0.05 \frac{{ }^{\circ} \mathrm{C} \mathrm{s}}{\mathrm{cal}}$$

Answer

$$0.02 \frac{{ }^{\circ} \mathrm{C} \mathrm{s}}{\mathrm{cal}}$$

Answer

$$5 \frac{{ }^{\circ} \mathrm{C} \mathrm{s}}{\mathrm{cal}}$$

Answer

$$2 \frac{{ }^{\circ} \mathrm{Cs}}{	ext { cal }}$$

Thermodynamics & KTG

The rate of flow of heat through a copper rod with temperature difference $28^{\circ} \mathrm{C} is 1400 \mathrm{~cal} \mathrm{~s}^{-1}$. The thermal resistance of copper rod will be

Choose the correct answer: