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A cup of coffee cools from 90C90^{\circ}\text{C} to 80C80^{\circ}\text{C} in tt minutes, when the room temperature is 20C20^{\circ}\text{C}. The time taken by a similar cup of coffee to cool from 80C80^{\circ}\text{C} to 60C60^{\circ}\text{C} at room temperature same at 20C20^{\circ}\text{C} is:

A

1013t\frac{10}{13}t

B

513t\frac{5}{13}t

C

1310t\frac{13}{10}t

D

135t\frac{13}{5}t

Step-by-Step Solution

According to Newton's Law of Cooling, the rate of cooling is proportional to the temperature difference between the body and the surroundings. For small temperature intervals, we use the average form: T1T2t=K[T1+T22Ts]\frac{T_1 - T_2}{t} = K \left[ \frac{T_1 + T_2}{2} - T_s \right]

Case 1: Cooling from 90C90^{\circ}\text{C} to 80C80^{\circ}\text{C} in time tt. 9080t=K[90+80220]\frac{90 - 80}{t} = K \left[ \frac{90 + 80}{2} - 20 \right] 10t=K\frac{10}{t} = K 10t=65K    K=1065t=213t\frac{10}{t} = 65K \implies K = \frac{10}{65t} = \frac{2}{13t}

Case 2: Cooling from 80C80^{\circ}\text{C} to 60C60^{\circ}\text{C} in time tt'. 8060t=K[80+60220]\frac{80 - 60}{t'} = K \left[ \frac{80 + 60}{2} - 20 \right] 20t=K\frac{20}{t'} = K 20t=50K\frac{20}{t'} = 50K

Substitute the value of KK from Case 1: 20t=50(213t)\frac{20}{t'} = 50 \left( \frac{2}{13t} \right) 20t=10013t\frac{20}{t'} = \frac{100}{13t} t=20×13t100=135tt' = \frac{20 \times 13t}{100} = \frac{13}{5}t

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