72−θ₀ In the second case,64−52 5=K[64+52 2−θ₀]or2.4=K58−θ₀ Dividing (i) by (ii), we get 3.2 2.4=72−θ₀ 58−θ₀or 8 6=72−θ₀ 58−θ₀or 4 3=72−θ₀ 58−θ₀ 4(58−θ₀)=3(72−θ₀) 232−4θ₀=216−3θ₀ 232−216=4θ₀−3θ₀ 16=θ₀ θ₀=16°C
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72−θ₀ In the second case,64−52 5=K[64+52 2−θ₀]or2.4=K58−θ₀ Dividing (i) by (ii), we get 3.2 2.4=72−θ₀ 58−θ₀or 8 6=72−θ₀ 58−θ₀or 4 3=72−θ₀ 58−θ₀ 4(58−θ₀)=3(72−θ₀) 232−4θ₀=216−3θ₀ 232−216=4θ₀−3θ₀ 16=θ₀ θ₀=16°C
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A body initially at 80°C cools to 64°C in 5 min and to 52°C in 10 min. The temperature of the surrounding is?
26°C
16°C
40°C
36°C
Solution:
According to Newton's law of cooling θ₁−θ₂ t=K[θ₁+θ₂ 2−θ₀] In the first case,80−64 5=K[80+64 2−θ₀]or3.2=K72−θ₀ In the second case,64−52 5=K[64+52 2−θ₀]or2.4=K58−θ₀ Dividing (i) by (ii), we get 3.2 2.4=72−θ₀ 58−θ₀or 8 6=72−θ₀ 58−θ₀or 4 3=72−θ₀ 58−θ₀ 4(58−θ₀)=3(72−θ₀) 232−4θ₀=216−3θ₀ 232−216=4θ₀−3θ₀ 16=θ₀ θ₀=16°C
