An object is cooled from 75^{circ}C to 65^{ci | vedclass

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(A) According to Newton's law of cooling,the rate of cooling is given by $\frac{dT}{dt} = K(T_{avg} - T_{surrounding})$.In the first case,the temperat

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$4$

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(A) According to Newton's law of cooling,the rate of cooling is given by $\frac{dT}{dt} = K(T_{avg} - T_{surrounding})$.In the first case,the temperature changes from $75^{\circ}C$ to $65^{\circ}C$ in $t_1 = 2$ minutes.$\frac{75 - 65}{2} = K \left( \frac{75 + 65}{2} - 30 \right)$$\frac{10}{2} = K(70 - 30) \implies 5 = 40K \implies K = \frac{5}{40} = \frac{1}{8}$ $(i)$In the second case,the temperature changes from $55^{\circ}C$ to $45^{\circ}C$ in $t_2$ minutes.$\frac{55 - 45}{t_2} = K \left( \frac{55 + 45}{2} - 30 \right)$$\frac{10}{t_2} = K(50 - 30) \implies \frac{10}{t_2} = 20K$ (ii)Dividing equation $(i)$ by (ii):$\frac{5}{10/t_2} = \frac{40K}{20K}$$\frac{5t_2}{10} = 2$$5t_2 = 20 \implies t_2 = 4$ minutes.

An object is cooled from $75^{\circ}C$ to $65^{\circ}C$ in $2$ minutes in a room at $30^{\circ}C$. The time taken to cool another object from $55^{\circ}C$ to $45^{\circ}C$ in the same room in minutes is

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