The variation of temperature in a plane wall is determined to be \(T(x)=110-60 x\) where \(x\) is in \(\mathrm{m}\) and \(T\) is in \({ }^{\circ} \mathrm{C}\). If the thickness of the wall is \(0.75 \mathrm{~m}\), the temperature difference between the inner and outer surfaces of the wall is (a) \(30^{\circ} \mathrm{C}\) (b) \(45^{\circ} \mathrm{C}\) (c) \(60^{\circ} \mathrm{C}\) (d) \(75^{\circ} \mathrm{C}\) (e) \(84^{\circ} \mathrm{C}\)

Short Answer

Expert verified
Answer: The temperature difference between the inner and outer surfaces of the wall is 45°C.

Step by step solution

01

Identify the values for the inner and outer surfaces.

The inner and outer surfaces of the wall are located at \(x = 0\) and \(x = 0.75\), respectively.
02

Calculate the temperature at the inner surface.

Substitute \(x = 0\) into the temperature function: \(T(0) = 110 - 60(0) = 110^\circ \mathrm{C}\).
03

Calculate the temperature at the outer surface.

Substitute \(x = 0.75\) into the temperature function: \(T(0.75) = 110 - 60(0.75) = 110 - 45 = 65^\circ \mathrm{C}\).
04

Calculate the temperature difference between the inner and outer surfaces.

Subtract the temperature at the outer surface from the temperature at the inner surface: Temperature difference = \(110^\circ \mathrm{C} - 65^\circ \mathrm{C} = 45^\circ\mathrm{C}\). The temperature difference between the inner and outer surfaces of the wall is (b) \(45^{\circ} \mathrm{C}\).

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