Concepts Of Physics MCQ Edition [Volume 2]
- Exam
- Concepts Of Physics MCQ Edition [Volume 2]
- Subject
- Physics
- Chapter
- Heat Transfer
Concepts Of Physics MCQ Edition [Volume 2] · Physics · Heat Transfer
A cubical box with a volume of 216 cm^3 is constructed from wood of...
A cubical box with a volume of is constructed from wood of thickness . Its interior is electrically heated using a heater. During steady state, the temperature difference between the inside and outside surfaces is observed to be 5^{}^{{}^{\circ}}\text{C}. Assuming all the consumed electrical energy is converted into heat, determine the thermal conductivity of the box's material.
Options
- A0.93 \text{ W m}^{-1} ^{}^{{}^{\circ}}\textC^{-1}
- B5.56 \text{ W m}^{-1} ^{}^{{}^{\circ}}\textC^{-1}
- C0.46 \text{ W m}^{-1} ^{}^{{}^{\circ}}\textC^{-1}
- D92.6 \text{ W m}^{-1} ^{}^{{}^{\circ}}\textC^{-1}
Correct answer
A. 0.93 \text{ W m}^{-1} ^{}^{{}^{\circ}}\textC^{-1}
Step-by-step solution
Volume of the cubical box is
Side length of the cubical box is
Total surface area of the cubical box is
Thickness of the wood is
In steady state, the rate of heat flow equals the power of the heater, so
Using the formula for thermal conduction:
Substituting the given values:
K = \dfrac{0.1}{0.108} \approx 0.926 \text{ W m}^{-1} ^{}^{{}^{\circ}}\textC^{-1}
Rounding to two decimal places, we get 0.93 \text{ W m}^{-1} ^{}^{{}^{\circ}}\textC^{-1}
Side length of the cubical box is
Total surface area of the cubical box is
Thickness of the wood is
In steady state, the rate of heat flow equals the power of the heater, so
Using the formula for thermal conduction:
Substituting the given values:
K = \dfrac{0.1}{0.108} \approx 0.926 \text{ W m}^{-1} ^{}^{{}^{\circ}}\textC^{-1}
Rounding to two decimal places, we get 0.93 \text{ W m}^{-1} ^{}^{{}^{\circ}}\textC^{-1}