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Grade 11Thermal Physics

The ratio of thermal conductivities 60Wm-1K-1, 440Wm-1K-1 and the areas 0.2m2,0.3m2, are connected in parallel to each other. The effective conductivity of the combination is....
a) 50Wm-1K-1 b) 45Wm-1K-1 c) 52Wm-1K-1 d) 48Wm-1K-1

Profile image of Sreenidhi Varadi
9 Years agoGrade 11
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1 Answer

Profile image of Askiitians Tutor Team
ApprovedApproved Tutor Answer0 Years ago

To determine the effective thermal conductivity of materials connected in parallel, we can use a specific formula that takes into account both the thermal conductivities and the areas of the materials involved. In this case, we have two materials with thermal conductivities of 60 W/m·K and 440 W/m·K, and areas of 0.2 m² and 0.3 m², respectively. Let's break down the calculation step by step.

Understanding the Concept of Parallel Conductivity

When materials are connected in parallel, the heat transfer through each material is additive. This means that the total heat transfer is the sum of the heat transfers through each individual material. The effective thermal conductivity (k_eff) can be calculated using the formula:

Formula for Effective Thermal Conductivity

The effective thermal conductivity for materials in parallel is given by:

k_eff = (k1 * A1 + k2 * A2) / (A1 + A2)

Where:

  • k1 = thermal conductivity of the first material
  • A1 = area of the first material
  • k2 = thermal conductivity of the second material
  • A2 = area of the second material

Plugging in the Values

Now, let's substitute the given values into the formula:

  • k1 = 60 W/m·K
  • A1 = 0.2 m²
  • k2 = 440 W/m·K
  • A2 = 0.3 m²

Now, we can calculate:

k_eff = (60 * 0.2 + 440 * 0.3) / (0.2 + 0.3)

Calculating Each Component

First, calculate the numerator:

  • 60 * 0.2 = 12
  • 440 * 0.3 = 132

Adding these gives:

12 + 132 = 144

Now, calculate the denominator:

0.2 + 0.3 = 0.5

Final Calculation

Now we can find the effective thermal conductivity:

k_eff = 144 / 0.5 = 288 W/m·K

Interpreting the Result

It seems that I made an error in interpreting the question. The effective conductivity calculated here is not among the options provided. Let's double-check the values and ensure we are interpreting the problem correctly.

Upon reviewing, it appears that the effective conductivity should be calculated based on the total heat transfer and the areas involved. If we consider the heat transfer through each material separately and then combine them, we can find the effective conductivity that matches one of the options.

However, based on the calculations provided, the effective thermal conductivity of the combination of the two materials in parallel is indeed 288 W/m·K, which does not match any of the options given. It may be worth revisiting the problem statement or the values provided to ensure accuracy.

In summary, the effective thermal conductivity for the given materials connected in parallel is calculated to be 288 W/m·K, which is significantly higher than the options listed. If you have any additional information or context, please share it, and we can refine our calculations further!