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A certain quantity of water cools from 70C70^{\circ}\text{C} to 60C60^{\circ}\text{C} in the first 5 minutes5\text{ minutes} and to 54C54^{\circ}\text{C} in the next 5 minutes5\text{ minutes}. The temperature of the surroundings will be:

A

45C45^{\circ}\text{C}

B

20C20^{\circ}\text{C}

C

42C42^{\circ}\text{C}

D

10C10^{\circ}\text{C}

Step-by-Step Solution

According to the average form of Newton's law of cooling: T1T2t=K(T1+T22Ts)\frac{T_1 - T_2}{t} = K \left( \frac{T_1 + T_2}{2} - T_s \right) where TsT_s is the temperature of the surroundings.

For the first 5 minutes5\text{ minutes}: 70605=K(70+602Ts)\frac{70 - 60}{5} = K \left( \frac{70 + 60}{2} - T_s \right) 2=K(65Ts)2 = K(65 - T_s) ... (i)

For the next 5 minutes5\text{ minutes}: 60545=K(60+542Ts)\frac{60 - 54}{5} = K \left( \frac{60 + 54}{2} - T_s \right) 1.2=K(57Ts)1.2 = K(57 - T_s) ... (ii)

Dividing equation (i) by (ii): 21.2=65Ts57Ts\frac{2}{1.2} = \frac{65 - T_s}{57 - T_s} 53=65Ts57Ts\frac{5}{3} = \frac{65 - T_s}{57 - T_s} 5(57Ts)=3(65Ts)5(57 - T_s) = 3(65 - T_s) 2855Ts=1953Ts285 - 5T_s = 195 - 3T_s 2Ts=902T_s = 90 Ts=45CT_s = 45^{\circ}\text{C}

Therefore, the temperature of the surroundings is 45C45^{\circ}\text{C}.

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