First slide

Heat transfer-Radiation

Question

A body cools from $70^{\circ} c$ to $50^{\circ} c$ in 6 minutes. If temperature of surroundings is $30^{\circ} c$ . What is the temperature after 12 minutes?

NA

A

$20^{\circ} c$
B

$30^{\circ} c$
C

$34^{\circ} c$
D

$40^{\circ} c$

Solution

$\begin{array}{rcl} \begin{array}{l} {initial temperature of the body is θ}_{1} {=70}^{o} c after 6 minutes \\ {temperature of body is θ}_{2} {=50}^{o} {c; surrounding temperature is θ}_{0} {=30}^{o} {c; final temperature of body is θ}_{} =? \\ {time t}_{1} =6 minutes \\ {time t}_{2} =12 minutes \\ f r o m n e w t o n l a w o f c o o l i n g \end{array} \\ \frac{dθ}{dt} =-k (\frac{θ_{1} {+θ}_{2}}{2} {-θ}_{0}) ---(1) \\ s u b s t i t u t e g i v e n v a l u e s i n e q n (1) \\ \frac{θ_{1} {-θ}_{2}}{t i m e} = - k (\frac{θ_{1} {+θ}_{2}}{2} {-θ}_{0}) \\ \frac{70 - 50}{6} & = & - k (\frac{70 + 50}{2} -30) \\ \frac{20}{6} & = & - k (60-30) \\ k & = & - \frac{1}{9} - - - (2) \\ s u b s t i t u t e s e c o n d p a r t o f q u e s t i o n, g i v e n v a l u e s a n d e q n (2) i n e q n (1) \\ \frac{50 - θ}{12} & = & - (- \frac{1}{9}) (\frac{50 + θ}{2} -30) \\ \Rightarrow & θ = 34^{0} C \end{array}$

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