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Thermal Physics

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Joined:	4 Nov 2011
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4 Dec 2011 12:04:07 IST

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Rate of cooling

Physics

A solid copper sphere cools at the rate of 3degree/min, when its temperature is 227 degree. Find the rate of cooling of another copper sphere of thrice its radius at a temperature of 427 degree. The room temperature is 27 degree in both cases.

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edison

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Joined: 19 Oct 2006

Posts: 7537

5 Dec 2011 14:15:54 IST

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Use Newtons law of cooling explained below:

Newton's Law of Cooling states that the hotter an object is, the faster it cools. More precisely, the rate of cooling is proportional to the temperature difference between an object and its surroundings. This word statement leads to the classic equation of exponential decline over time that applies to many phenomena in science and engineering, including the discharge of a capacitor and the decay in radioactivity.

Newton's Law of Cooling is useful when studying water heating because it can tell us how fast the hot water in pipes cools off, and also tells us how fast a water heater cools down if you turn off the breaker when you go on vacation.

The basic equation for Newton's Law of Cooling is:

T(t) = T_A + (T_H-T_A) e^-kt... (1)

where
T(t) = Temperature at time t
T_A = Ambient temperature (temp of surroundings)
T_H = Temperature of hot object at time 0
k = positive constant
t = time

Now in (1) substitute, t = 1s, T(t) = 273 + (227 - 3) = 497 K

T_A = RT = 300 K

T_H = 273 227 = 500 K

Now find the value of constant 'k'.

Having determined 'k' work out for the rate of cooling for copper sphere at temperature 427 + 273 = 700 K

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