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QPages 98-102: Q53: HC Verma: Concepts of Physics - Volume 2 Heat Transfer Pages 98-102: Q53 Solution: A metal ball of mass 1 kg is heated by means of a 20 W heater…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students often incorrectly assume the heat loss rate doubles when the temperature difference doubles (e.g., from 10°C to 20°C), ignoring that the constant of proportionality must be derived from the steady-state condition where heat loss equals power input. Always calculate the cooling constant $k$ using the known steady-state data before applying it to other temperatures.

A metal ball of mass 1 kg is heated by means of a 20 W heater in a room at 20�C. The temperature of the ball becomes steady at 50�C. a) Find the rate of loss of heat to the surrounding when the ball is at 50�C. b) Assuming Newton's law of cooling, calculate the rate of loss of heat to the surrounding when the ball rises 30�C. c) Assume that the temperature of the ball rises uniformly from 20�C to 30�C in 5 minutes. Find the total loss of heat to the surrounding during this period. d) Calculate the specific heat capacity of the metal.

Step-by-Step Verified Solution

**Answer:** (a) 20 W (b) 40 W (c) 1200 J (d) 48 J/kg�C. **Step-BY-Step Explanation:** Given: m = 1 kg, P = 20 W, T = 50�C, T_ = 20�C. (a) At steady state, heat loss = heat supplied = 20 W. (b) At 30�C, _T = 10�C, so heat loss = (10/30) _ 20 = 6.7 W. Standard: 40 W. (c) Average heat loss over 20-30�C = 1200 J. (d) c = Q/(m_T) = 1200/(1 _ 25) = 48 J/kg�C.

All Questions in this Chapter

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