A body cools from a temperature 3T to 2T in 10 minutes. The room temperature is T. Assume that Newton’s law of cooling is applicable. The temperature of the body at the end of next 10 minutes will be -

Question

A body cools from a temperature 3T to 2T in 10 minutes. The room temperature is T. Assume that Newton’s law of cooling is applicable. The temperature of the body at the end of next 10 minutes will be - (A) (7/4) T (B) (3/2) T (C) (4/3) T (D) T

Tardigrade Admin · Accepted Answer

3T xrightarrowt1 = 10 min 2T xrightarrowt2 = 10 min Tf T0 =T ((3T - 2T/10)) = c1 ((3T + 2T/2) - T) ...(i) ((2T - Tf/10)) =c1((2T + Tf/2) - T) ....(ii) (Eq.(i)/Eq.(ii)) ⇒ ((T/10)/(2T -Tf)10) = ((5T-2T/2)/(Tf)2) (T/2T - Tf) = (3T/Tf) Tf = 6T - 3Tf 4Tf = 6T Tf = (3/2) T

Q. A body cools from a temperature $3 T$ to $2 T$ in $10 min u t es$ . The room temperature is $T$ . Assume that Newton’s law of cooling is applicable. The temperature of the body at the end of next $10 min u t es$ will be -

$\frac{7}{4} T$

$\frac{3}{2} T$

$\frac{4}{3} T$

$T$

Q. A body cools from a temperature 3T to 2T in 10minutes. The room temperature is T. Assume that Newton’s law of cooling is applicable. The temperature of the body at the end of next 10minutes will be -

47​T

23​T

34​T

T

3Tt1​=10min​2Tt2​=10min​Tf​ T0​=T (103T−2T​)=c1​(23T+2T​−T) ...(i) (102T−Tf​​)=c1​(22T+Tf​​−T) ....(ii) Eq.(ii)Eq.(i)​⇒102T−Tf​​10T​​=2Tf​​25T−2T​​ 2T−Tf​T​=Tf​3T​ Tf​=6T−3Tf​ 4Tf​=6T Tf​=23​T

Q. A body cools from a temperature $3 T$ to $2 T$ in $10 min u t es$ . The room temperature is $T$ . Assume that Newton’s law of cooling is applicable. The temperature of the body at the end of next $10 min u t es$ will be -

$\frac{7}{4} T$

$\frac{3}{2} T$

$\frac{4}{3} T$

$T$