Temperature of hot soup in a bowl goes 98∘C to 86∘C in 2 min. The temperature of surrounding is 22∘C. Find the time taken for the temperature of soup to go from 75∘C to 69∘C. (Assume Newton’s Law of cooling is valid)

We have , ΔθΔt=−K(θ1+θ22−θ0)Given θ0=220C98−862=−K(98+862−22)........(1)75−69t2=−K(75+692−22)..........(2)From (1) and (2)t2=7050=1.4 min

Temperature of hot soup in a bowl goes 98∘C to 86∘C in 2 min. The temperature of surrounding is 22∘C. Find the time taken for the temperature of soup to go from 75∘C to 69∘C. (Assume Newton’s Law of cooling is valid)

We have , ΔθΔt=−K(θ1+θ22−θ0)Given θ0=220C98−862=−K(98+862−22)........(1)75−69t2=−K(75+692−22)..........(2)From (1) and (2)t2=7050=1.4 min

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Temperature of hot soup in a bowl goes $98^{\circ} C$ to $86^{\circ} C$ in $2 m i n$ . The temperature of surrounding is $22^{\circ} C$ . Find the time taken for the temperature of soup to go from $75^{\circ} C$ to $69^{\circ} C$ . (Assume Newton’s Law of cooling is valid)

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2 min

1.4 min

3.2 min

1 min

answer is B.

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Detailed Solution

We have , $\frac{Δ θ}{Δ t} = - K (\frac{θ_{1} + θ_{2}}{2} - θ_{0})$
Given $θ_{0} = 22^{0} C$
$\frac{98 - 86}{2} = - K (\frac{98 + 86}{2} - 22) ........ (1)$
$\frac{75 - 69}{t_{2}} = - K (\frac{75 + 69}{2} - 22) .......... (2)$
From (1) and (2)
$t_{2} = \frac{70}{50} = 1.4 \min$