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Newton’s Law of Cooling states that the rate of temperature of the body is proportional to the difference between the temperature of the body and that of the surrounding medium. It is a reasonably accurate approximation in some circumstances.
More precisely, let T denote the temperature of an object and Tο the ambient temperature. if t denotes time, then Newton's law states that
dT/dt = -k(T-Tο)
where k is a positive constant. Thus, if the object temperature is much higher than its surroundings, then T-Tο is large and positive, so dTdT/dt is large and negative, so the object cools quickly. If the object temperature is only slightly higher than its surroundings, then T-Tο is small positive, and the object cools slowly. So a cup of coffee will cool more quickly if you put it in the refrigerator!
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Newton’s Law of Cooling states that the rate of change of temperature of a body is proportional to the difference between the temperature of the body and that of the surrounding medium. It is a reasonably accurate approximation in some circumstances.
More precisely, let T denote the temperature of an object and Tο the ambient temperature. if t denotes time, then Newton's law states that
dT/dt = -k(T-Tο)
where k is a positive constant. Thus, if the object temperature is much higher than its surroundings, then T-Tο is large and positive, so dTdT/dt is large and negative, so the object cools quickly. If the object temperature is only slightly higher than its surroundings, then T-Tο is small positive, and the object cools slowly. So a cup of coffee will cool more quickly if you put it in the refrigerator!
read less
It states that rate of cooling of a body is proportional to the excess temperature of body over surroundings. dQ/dt= -K(T2-T1) where T1 is temperature of surroundings T2 is temperature of body. It is our observation that cloths get dry early when outside temperature is more than in winter. Heat lost by body is -msdT2 when temperature falls by small amount T2 in small time dt. m mass of body s is specific heat capacity.T2 is temp of body.
Rate of loss of heat dQ/dt=msdT2/dt
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