A hot body placed in air cools down to a lowe | vedclass

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(D) According to Newton's law of cooling, the rate of change of temperature is given by $\frac{dT}{dt} = -k(T - T_0)$, where $T_0$ is the surrounding

Vedclass · Accepted Answer

$\frac{\ln (5)}{\ln (2.5)}$

Vedclass · Answer

(D) According to Newton's law of cooling, the rate of change of temperature is given by $\frac{dT}{dt} = -k(T - T_0)$, where $T_0$ is the surrounding temperature.Integrating this, we get $T(t) - T_0 = (T_i - T_0)e^{-kt}$, where $T_i$ is the initial temperature.The heat lost by the body is $Q(t) = mc(T_i - T(t))$. The maximum heat it can lose is $Q_{max} = mc(T_i - T_0)$.Thus, the fraction of heat lost is $\frac{Q(t)}{Q_{max}} = \frac{T_i - T(t)}{T_i - T_0} = 1 - e^{-kt}$.For time $t_1$, the body loses $60 \%$ of maximum heat: $0.6 = 1 - e^{-kt_1} \Rightarrow e^{-kt_1} = 0.4 \Rightarrow t_1 = \frac{1}{k} \ln(\frac{1}{0.4}) = \frac{1}{k} \ln(2.5)$.For time $t_2$, the body loses $80 \%$ of maximum heat: $0.8 = 1 - e^{-kt_2} \Rightarrow e^{-kt_2} = 0.2 \Rightarrow t_2 = \frac{1}{k} \ln(\frac{1}{0.2}) = \frac{1}{k} \ln(5)$.The ratio $\frac{t_2}{t_1} = \frac{\ln(5)}{\ln(2.5)} = \frac{\ln(5)}{\ln(5/2)}$.

$A$ hot body placed in air cools down to a lower temperature. The rate of decrease of temperature is proportional to the temperature difference from the surrounding. The body loses $60 \%$ and $80 \%$ of the maximum heat it can lose in time $t_1$ and $t_2$ respectively. The ratio $t_2 / t_1$ will be

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