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Certain quantity of water cools from 70°C to 60°C in the first 5 minutes and to 54°C in the next 5 minutes. The temperature of the surroundings is
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Correct option is A
Let Ts be the temperature of the surroundings. According to Newton's law of coolingT1−T2t=KT1+T22−Tsfor first 5 minutes,T1=70∘C,T2=60∘C,t=5 minutes ∴70−605=K70+602−Ts105=K65−TsT1=60∘C,T2=54∘C,t=5 minutes ∴ 60−545=K60+542−Ts65=K57−TsDivide eqn. (i) by eqn. (ii), we get53=65−Ts57−Ts285−5Ts=195−3Ts2Ts=90 or Ts=45∘CSimilar Questions
A calorimeter of mass 0.2 kg and specific heat 900 Containing 0.5 kg of a liquid of specific heat 2400J/kg/K. Its temperature falls from
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