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A liquid in beaker has temperature theta...

A liquid in beaker has temperature `theta(t)` at time t and `q_0` is temperature of surroundings, then according to Newton's law of cooling, the correct graph between `log_e(theta-theta_0)` and t is

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Knowledge Check

  • If a piece of metal is heated to temperature theta and then allowed to cool in a room which is at temperature 0_0 , the graph between the temperature T of the metal and time t will be closest to

    A
    B
    C
    D

  • According to Stefan"s Law, heat energy emitted//sec//area by a specific black body varies directly as the fourth power of its absolute temperature. The wavelength corresponding to which energy emitted is maximum varies inversely as the temperature of black body (Wein's Law). However, the rate of loss of heat of a liquid varies directly as the difference in temperatures of the liquid and the surroundings, provided this difference is small (~~30^@C) This is Newton's law of cooling Answer the following question based on above passage : A liquid takes 5 min to cool from 60^@C to 50^@C , when temperature of surroundings is 30^@C . How long will it take to cool from 50^@C to 40^@C ?

    A
    5 minutes
    B
    4 minutes
    C
    6 minutes
    D
    10 minutes

  • According to Stefan"s Law, heat energy emitted//sec//area by a specific black body varies directly as the fourth power of its absolute temperature. The wavelength corresponding to which energy emitted is maximum varies inversely as the temperature of black body (Wein's Law). However, the rate of loss of heat of a liquid varies directly as the difference in temperatures of the liquid and the surroundings, provided this difference is small (~~30^@C) This is Newton's law of cooling Answer the following question based on above passage : A liquid takes 5 min to cool from 60^@C to 50^@C , when temperature of surroundings is 30^@C . If surrounding temp. reduces to 10^@C , time taken to cool from 60^@C50^@C would be

    A
    5 min
    B
    4 min
    C
    6 min
    D
    7 min

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