# Find thermal conductivity of metal in a rod

• confused1
In summary, the thermal conductivity of the metal in the rod is 320 W/mK and the oven temperature needed to maintain a heat current of 50W is 1444.69 K.
confused1
A cylinder with a piston is filled with ideal gas. The gas temperature is held at 500K. The cylinder is heated by an oven through a square metal rod connected between the oven and the cylinder. The rod has sides 2.5cm and length 2m.

(a) If the oven is held at 1300K, and the heat current conducted through the metal rod is 20W, find the thermal conductivity of the metal in the rod.

HELP: H = k*A*(T2-T1) / L

HELP: H is the heat current; k is thermal conductivity; T2,T1 are temperatures at two ends; A, L are cross-section area and length of the rod.

(b) In order to maintain the temperature of the cylinder, a heat current of 50W is needed. If we use the same metal rod to conduct heat, find the oven temperature in Kelvin that will maintain the necessary heat current.

HELP: H = k*A*(T2-T1) / L Answer:(a) The thermal conductivity of the metal in the rod is calculated by rearranging the equation given as follows: k = H * L / (A * (T2 - T1)) = 20 W * 2m / (0.00625 m^2 * (1300 K - 500 K)) = 320 W/mK (b) The oven temperature can be calculated using the same equation as follows: T2 = T1 + (H*L)/(k*A) = 500 K + (50 W * 2m) / (320 W/mK * 0.00625 m^2) = 1444.69 K

(a) To find the thermal conductivity of the metal in the rod, we can use the formula provided: H = k*A*(T2-T1)/L. We are given the heat current (H) of 20W, the length (L) of the rod as 2m, and the temperature difference (T2-T1) as 1300K-500K = 800K. The area (A) of the rod can be calculated as 2.5cm*2.5cm = 6.25cm^2 or 0.000625m^2. Plugging these values into the formula, we get: 20W = k*0.000625m^2*800K/2m. Solving for k, we get a thermal conductivity of approximately 0.02 W/mK. Therefore, the thermal conductivity of the metal in the rod is 0.02 W/mK.

(b) To find the oven temperature needed to maintain a heat current of 50W, we can rearrange the formula to solve for T2-T1: T2-T1 = H*L/(k*A). Again, we are given the length (L) of the rod as 2m, the heat current (H) as 50W, and the thermal conductivity (k) as 0.02 W/mK. The area (A) of the rod remains the same at 0.000625m^2. Plugging these values into the formula, we get: T2-T1 = 50W*2m/(0.02 W/mK*0.000625m^2) = 160000K. Therefore, the oven temperature needed to maintain a heat current of 50W is 160000K + 500K = 160500K or approximately 160227°C.

## 1. What is thermal conductivity?

Thermal conductivity is the measure of a material's ability to conduct heat. It is a property that determines how well a material can transfer heat from one point to another.

## 2. Why is it important to find the thermal conductivity of metal in a rod?

Knowing the thermal conductivity of a metal rod is important in various applications, such as designing heat transfer systems, understanding thermal insulation, and predicting the behavior of materials under different temperature conditions.

## 3. How is the thermal conductivity of a metal rod measured?

The thermal conductivity of a metal rod is typically measured using a device called a thermal conductivity meter or a heat flow meter. This device measures the amount of heat flowing through the rod and calculates the thermal conductivity based on the geometry and temperature difference of the rod.

## 4. What factors affect the thermal conductivity of a metal rod?

The thermal conductivity of a metal rod is affected by various factors, including the type of metal, temperature, and microstructure of the metal. Other factors that can influence thermal conductivity include impurities, defects, and surface roughness of the rod.

## 5. How can the thermal conductivity of a metal rod be improved?

The thermal conductivity of a metal rod can be improved by using materials with higher thermal conductivity, optimizing the microstructure of the metal, and reducing impurities and defects in the metal. Additionally, using thermal insulation or heat transfer coatings on the rod can also help improve its thermal conductivity.

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