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Physics Question on thermal properties of matter

A hot liquid is filled in a container and kept in a room of temperature of 25∘^{\circ}C. The liquid emits heat at the rate of 200 Js−1^{-1} when its temperature is 75∘^{\circ}C. When the temperature of the liquid becomes 40∘^{\circ}C, the rate of heat loss is J s−1^{-1} is

A

160

B

140

C

80

D

60

Answer

60

Explanation

Solution

Rate of cooling (−dθdt)=α(θ−θ0) \bigg(\frac{-d\theta}{dt}\bigg)=\alpha (\theta-\theta_0)
or           (−dθdt)1(−dθdt)2=θ1−θ0θ2−θ0or \, \, \, \, \, \, \, \, \, \, \, \frac{\bigg(\frac{-d\theta}{dt}\bigg)_1}{\bigg(\frac{-d\theta}{dt}\bigg)_2}=\frac{\theta_1 -\theta_0}{\theta_2 -\theta_0}
or           200(−dθdt)2=75−2540−25or \, \, \, \, \, \, \, \, \, \, \, \frac{200}{\bigg(\frac{-d\theta}{dt}\bigg)_2} =\frac{75-25}{40-25}
or       (−dθdt)2=1520×200=60Js−1or \, \, \, \, \, \, \, \bigg(-\frac{d\theta}{dt}\bigg)_2 =\frac{15}{20} \times 200 =60Js^{-1}