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 the end of next 10 minutes will bedfrac-7-4-Tdfrac-4-3-Tdfrac-3-2-TT

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From newton-apos-s law of cooling-ln-3T-x2212-T2T-x2212-T-kt-ln-2-k-10-For next 10 minutes-ln-2T-x2212-TT-x2032-x2212-T-k-10-x27F9-2T-x2212-TT-x2032-x2212-T-2-x27F9-T-x2032-32T

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
$\frac{7}{4} T$
$\frac{4}{3} T$
$\frac{3}{2} T$
$T$

From newton's law of cooling,
$l n (\frac{3 T - T}{2 T - T}) = k t$

$= l n (2) = k (10)$

For next 10 minutes,

$l n (\frac{2 T - T}{T^{'} - T}) = k (10)$

$⟹ \frac{2 T - T}{T^{'} - T} = 2$

$⟹ T^{'} = \frac{3}{2} T$

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 be74T43T32TT

From newton's law of cooling,ln(3T−T2T−T)=kt=ln(2)=k(10)For next 10 minutes,ln(2T−TT′−T)=k(10)⟹2T−TT′−T=2⟹T′=32T

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
$\frac{7}{4} T$
$\frac{4}{3} T$
$\frac{3}{2} T$
$T$

From newton's law of cooling,
$l n (\frac{3 T - T}{2 T - T}) = k t$

$= l n (2) = k (10)$

For next 10 minutes,

$l n (\frac{2 T - T}{T^{'} - T}) = k (10)$

$⟹ \frac{2 T - T}{T^{'} - T} = 2$

$⟹ T^{'} = \frac{3}{2} T$