- Heat Transfer - Section 1
- Heat Transfer - Section 2
- Heat Transfer - Section 3
- Heat Transfer - Section 4
- Heat Transfer - Section 5
- Heat Transfer - Section 6
- Heat Transfer - Section 7
- Heat Transfer - Section 8
- Heat Transfer - Section 9
- Heat Transfer - Section 10
- Heat Transfer - Section 11


Heat Transfer - Engineering
Q1: A metal ball of radius 0.1 m at a uniform temperature of 90°C is left in air at 30°C. The density and the specific heat of the metal are 3000 kg/m3 and 0.4 kJ/kg.K respectively. The heat transfer co-efficient is 50 W/m2.K Neglecting the temperature gradients inside the ball, the time taken (in hours) for the ball to cool to 60°C isA 555
B 55.5
C 0.55
D 0.15
ANS:D - 0.15 To find the time taken for the metal ball to cool from 90°C to 60°C, we can use Newton's law of cooling, which states that the rate of heat loss of a body is directly proportional to the temperature difference between the body and its surroundings. The rate of heat loss can be expressed as: =ℎ(−ambient)dtdQ=hA(T−Tambient) Where:
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