A body cool from 70^ C to 40^ C in 5 min. Calculate the time it takes to cool from 60^ C to 40^ C. The temperature of the surrounding is 20^ C.

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Accepted Answer

Correct answer: D. (d) Given, temperature of surrounding, T_s=20^ C. We know by Newton's law of looking T_i-T_ft=K( T_i+T_f2-T_s) … … (i ) According to first case, By using Eq. (i), we get T_1-T_2t_1=K( T_1+T_22-T_s) 70-405=K( 70+402-20)( ∵T_1=70^ CT_2=40^ C) t=15 sec 305=K( 1102-20) ⇒6=K (35 ) K= 635 According to second case, again by using Eq. (i), we get T_3-T_4T_2=K( T_3+T_42-T_s) ⇒ 60-40t_2= 635( 60+402-20) ⇒ ( ∵T_3=60^ C ,T_4=40^ C)t_2= 635 30 ⇒ 20t_2= 67 6 ⇒t_2= 20 736= 14036=3.89 min

$A body cool from 70^{\circ} C to 40^{\circ} C in 5 min . Calculate the time it takes to cool from 60^{\circ} C to 40^{\circ} C . The temperature of the surrounding is 20^{\circ} C .$

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