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Range of Quadratic Expression

Hot water coo...

Question

Hot water cools from $60^{o} C$ to $50^{o} C$ in the first $10$ minutes and to $42^{o} C$ in the next $10$ minutes. The temperature of the surroundings is :

25^{o} C

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10^{o} C

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15^{o} C

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20^{o} C

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Solution

The correct option is B $10^{o} C$
Answer is B
We know from Newton's law of cooling that .
$T = T_{f} + (T_{i} - T_{f}) e^{- k t}$
$50 = T_{f} + (60 - T_{f}) e^{- 10 k}$
$50 = y + 60 x$ _______________(i)
$w h e r e x = e^{- 10 k}$
and $y = T_{f} (1 - e^{- 10 k})$
Also,
$42 = T_{f} + (50 - T_{f}) e^{- 10 k}$
$42 = y + 50 x$ ______________ (ii)
solving (i) and (ii), we get
$x = e^{- 10 k} = 4 / 5$
and,
$y = T_{f} (1 - e^{- 10 k}) = 2$
$T_{f} = 10^{o} C$

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