If the surrounding air is kept at 25^{circ} C | vedclass

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(D) According to Newton's law of cooling,the rate of change of temperature is proportional to the temperature difference between the body and its surr

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$36.36^{\circ} C$ approximately

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(D) According to Newton's law of cooling,the rate of change of temperature is proportional to the temperature difference between the body and its surroundings:$\frac{d	heta}{dt} = -k(	heta - 	heta_0)$Integrating this,we get $\ln(	heta - 	heta_0) = -kt + C$.Given $	heta_0 = 25^{\circ} C$. At $t = 0$,$	heta = 80^{\circ} C$,so $\ln(80 - 25) = C \Rightarrow C = \ln(55)$.At $t = 30$,$	heta = 50^{\circ} C$,so $\ln(50 - 25) = -k(30) + \ln(55)$.$-30k = \ln(25) - \ln(55) = \ln(\frac{25}{55}) = \ln(\frac{5}{11})$.So,$k = -\frac{1}{30} \ln(\frac{5}{11}) = \frac{1}{30} \ln(\frac{11}{5})$.Now,for $t = 60 	ext{ minutes}$,$\ln(	heta - 25) = -k(60) + \ln(55)$.$\ln(	heta - 25) = -(\frac{1}{30} \ln(\frac{11}{5}))(60) + \ln(55) = -2 \ln(\frac{11}{5}) + \ln(55) = \ln((\frac{5}{11})^2) + \ln(55) = \ln(\frac{25}{121} 	imes 55) = \ln(\frac{25 	imes 5}{11}) = \ln(\frac{125}{11})$.$	heta - 25 = \frac{125}{11} \approx 11.36$.$	heta = 25 + 11.36 = 36.36^{\circ} C$.

If the surrounding air is kept at $25^{\circ} C$ and a body cools from $80^{\circ} C$ to $50^{\circ} C$ in $30 \text{ minutes}$,then the temperature of the body after one hour will be

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