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Q.A cup of coffee at temperature 100°C100°C is placed in a room whose temperature is 15°C15°C and it cools to 60°C60°C in 5 minutes. Find its temperature after a further interval of 5 minutes.

Tamil Nadu DgeTamil Nadu HSC (DGE) Board 2017Subjective· 10mImportance★★★★★
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Solve Newton's law of cooling as a linear first-order ODE, fit the constant using the given data point, then evaluate at t=10t=10.

  1. Set up the differential equation. Newton's law of cooling states the rate of cooling is proportional to the excess of temperature over the surroundings (15°C15°C):

    dTdt=−k(T−15),k>0.\frac{dT}{dt} = -k(T-15), \qquad k>0.

  2. Solve the ODE (variables separable).

    ∫dTT−15=−∫k dt  ⟹  ln⁡(T−15)=−kt+C1  ⟹  T−15=Ce−kt.\int \frac{dT}{T-15} = -\int k\,dt \implies \ln(T-15) = -kt + C_1 \implies T-15 = Ce^{-kt}.

  3. Apply the initial condition T(0)=100T(0)=100.

    100−15=Ce0  ⟹  C=85.SoT−15=85e−kt.100-15 = Ce^{0} \implies C=85. \quad\text{So}\quad T-15 = 85e^{-kt}.

  4. Apply the given data T(5)=60T(5)=60 to find e−5ke^{-5k}.

    60−15=85e−5k  ⟹  45=85e−5k  ⟹  e−5k=4585=917.60-15 = 85e^{-5k} \implies 45 = 85e^{-5k} \implies e^{-5k} = \frac{45}{85} = \frac{9}{17}.

  5. Find TT at t=10t=10 (a further 5 minutes after t=5t=5, so total t=10t=10): …

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