Question

The thickness of a uniform rectangular metal plate is 5 mm and the area of each surface is 3750 cm\(^2\). In steady state, the temperature difference between the two surfaces of the plate is 14°C. If the heat flowing through the plate in one second from one surface to the other surface is 42 J, then the thermal conductivity of the metal is:

• A. \( 90 \, \text{Wm}^{-1}\text{K}^{-1} \)
• B. \( 30 \, \text{Wm}^{-1}\text{K}^{-1} \)
• C. \( 45 \, \text{Wm}^{-1}\text{K}^{-1} \)
• D. \( 60 \, \text{Wm}^{-1}\text{K}^{-1} \) \bigskip

Hint:

To determine the final temperature, we apply the principle of heat exchange: - Heat lost by warm water is used to melt the ice and raise its temperature. - Latent heat of fusion of ice (334 J/g) is considered in calculations. - After melting, the total heat balance equation gives the final temperature. Understanding heat transfer and phase changes is crucial in thermodynamics.

Answer 1

The Correct Option is B

The heat transfer through the plate is governed by Fourier's law of heat conduction: \[ Q = \frac{k A \Delta T}{d} \] Where: - \( Q \) is the heat transferred per second (42 J), - \( k \) is the thermal conductivity (which we need to find), - \( A \) is the area of the cross-section of the plate (3750 cm\(^2\) = \( 3750 \times 10^{-4} \) m\(^2\)), - \( \Delta T \) is the temperature difference between the two surfaces (14°C), - \( d \) is the thickness of the plate (5 mm = 5 × 10\(^{-3}\) m). Rearranging the equation to solve for \( k \): \[ k = \frac{Q \cdot d}{A \cdot \Delta T} \] Substitute the known values: \[ k = \frac{42 \cdot (5 \times 10^{-3})}{3750 \times 10^{-4} \cdot 14} \] Thus, the thermal conductivity of the metal is \( 30 \, \text{Wm}^{-1}\text{K}^{-1} \).