A body cools from a temperature 3 T to 2 T 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 3 T to 2 T 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

(T1 - T2/Δ t)=K((T1 + T2/2) - T0) (3 T - 2 T/10)=K(2.5 T - T) ⇒ (T/10)=K(1.5)T K=(1/15) Now, (2 T - x/10)=K((2 T + x/2) - T) Solving x=(3 T/2)

Q. A body cools from a temperature $3 T$ to $2 T$ 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

$\frac{7}{4} T$

$\frac{3}{2} T$

$\frac{4}{3} T$

$T$

$\frac{T _{1} - T _{2}}{Δ t} = K (\frac{T _{1} + T _{2}}{2} - T_{0})$
$\frac{3 T - 2 T}{10} = K (2.5 T - T)$
$\Rightarrow \frac{T}{10} = K (1.5) T$
$K = \frac{1}{15}$
Now, $\frac{2 T - x}{10} = K (\frac{2 T + x}{2} - T)$
Solving $x = \frac{3 T}{2}$

Q. 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

47​T

23​T

34​T

T

ΔtT1​−T2​​=K(2T1​+T2​​−T0​) 103T−2T​=K(2.5T−T) ⇒10T​=K(1.5)T K=151​ Now, 102T−x​=K(22T+x​−T) Solving x=23T​

Q. A body cools from a temperature $3 T$ to $2 T$ 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

$\frac{7}{4} T$

$\frac{3}{2} T$

$\frac{4}{3} T$

$T$

$\frac{T _{1} - T _{2}}{Δ t} = K (\frac{T _{1} + T _{2}}{2} - T_{0})$
$\frac{3 T - 2 T}{10} = K (2.5 T - T)$
$\Rightarrow \frac{T}{10} = K (1.5) T$
$K = \frac{1}{15}$
Now, $\frac{2 T - x}{10} = K (\frac{2 T + x}{2} - T)$
Solving $x = \frac{3 T}{2}$