Hot water cools from $60^{0} C$ to $50^{0} C$ in the first $10$ minutes and to $42^{0} C$ in the next $10$ minutes. Then the temperature of the surroundings is

15^{0} C

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10^{0} C

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20^{0} C

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30^{0} C

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Solution

The correct option is B $10^{0} C$
By newton's law of cooling;
$\frac{\frac{10}{10}}{\frac{8}{10}} = \frac{e σ A {T_{o}}^{4} (55 - T_{o})}{e σ A {T_{o}}^{4} (46 - T_{o})}$
where 55 is average of 50 and 60, 46 is average of 42 and 50
$\Rightarrow \frac{10}{8} = \frac{55 - T_{o}}{46 - T_{o}} \Rightarrow 460 - 10 T_{o} = 440 - 8 T_{o} \Rightarrow 10 T_{o} - 8 T_{o} = 460 - 440 \Rightarrow 2 T_{o} = 20 \Rightarrow T_{o} = 10 ℃$

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The correct option is B 100CBy newton's law of cooling;1010810=eσATo4(55−To)eσATo4(46−To)where 55 is average of 50 and 60, 46 is average of 42 and 50⇒108=55−To46−To⇒460−10To=440−8To⇒10To−8To=460−440⇒2To=20⇒To=10℃

Hot water cools from $60^{0} C$ to $50^{0} C$ in the first $10$ minutes and to $42^{0} C$ in the next $10$ minutes. Then the temperature of the surroundings is