HC Verma: Concepts of Physics - Volume 2 Heat Transfer Pages 98-102: Q54 Solution: A metal block of heat capacity placed in a room at 20�C is…

QPages 98-102: Q54: HC Verma: Concepts of Physics - Volume 2 Heat Transfer Pages 98-102: Q54 Solution: A metal block of heat capacity placed in a room at 20�C is…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students often take the temperature‑rise rate (2 °C s⁻¹) and treat it as if it obeys Newton's cooling law, using it to compute the cooling constant. The correct constant k must be obtained from the *cooling* rate (0.2 °C s⁻¹) after the heater is off; then power loss at any temperature follows $P=k(T-T_{room})$. Using the heating rate leads to an over‑estimate of radiated power.

A metal block of heat capacity

80 \text{ J �C}^{-1}

placed in a room at 20�C is heated electrically. The heater is switched off when the temperature reaches 30�C. The temperature of the block rises at the rate of

2 \text{ �C s}^{-1}

just after the heater is switched on and falls at the rate of

0.2 \text{ �C s}^{-1}

just after the heater is switched off. Assume Newton's law of cooling to hold. 1. Find the power of the heater. 2. Find the power radiated by the block just after the heater is switched off. 3. Find the power radiated by the block when the temperature of the block is 25�C. 4. Assuming that the power radiated at 25�C represents the average value in the heating process, find the time for which the heater was kept on.

Step-by-Step Verified Solution

**Answer:** (1) 160 W (2) 16 W (3) 8 W (4) 4 s. **Step-BY-Step Explanation:** Given: Heat capacity = 80 J/�C, room temp = 20�C, T = 30�C. (1) Heating rate = 2�C/s _ 80 J/�C = 160 W. (2) Cooling rate = 0.2�C/s _ 80 = 16 W. (3) At 25�C: (25_20)/(30_20) _ 16 = 8 W. (4) Time = (80 _ 10)/(160_average loss) � 4 s.

QPages 98-102: Q54: HC Verma: Concepts of Physics - Volume 2 Heat Transfer Pages 98-102: Q54 Solution: A metal block of heat capacity placed in a room at 20�C is…

Step-by-Step Verified Solution

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