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State Nweton's law of net loss of heat. Hence, show that (dtheta)/(dt) prop(theta-theta_(@)). |
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Answer» Solution :Newton's law of cooling : The rate of loss of heat by a BODY is directly proportional to its excess of temperature over the surroundings, provided the excess is small. Let m, c and `theta` be the mass, speciffic heat capacity and temperature of a body RESPECTIVELY and let `theta_(@)`, assumed to be less than `theta`, be the temperature of the surroundings. As `theta gt theta_(@)`, the body loses heat to the surroundings. ACCORDINGS to Nweton's law of cooling, the rate of loss of heat by the body, `(dQ)/(dt) PROP(theta-theta_(@)` `:. (dQ)/(dt)=K(theta-theta_(@))` where K is a constant that depends on the body and the surroundings. But `(dQ)/(dt)=mc(dtheta)/(dt)` where `dtheta//dt` is the rate of fall of temperature of the body. ltBRgt `:. MC(DTHETA)/(DT)=k(theta-theta_(@)) :. (dtheta)/(dt)=(K)/(mc)(theta-theta_(@))` `:. (dtheta)/(dt)=k(theta-theta_(@))`, where `k=K//mc` is a constant. `:.` Rate of cooling of the body, `(dtheta)/(dt) prop(theta-theta_( |
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