Suppose we take 100 ml of water which has a temperature of 398 k and leave it in a room where the temperature is 298 k .what will happen?
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What will happen if we take 100ml of water which has a temperature of 378k and leave it in a room where the temperature is 298k?
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Srinidhi Kulkarni, Physics enthusiast
Answered Sep 24, 2016
Qualitatively answering, it's temperature will decrease with time from 378k to 298k in an exponential manner. This is because the evolution of temperature is given by
mC(dT/dt) = -hA(T-298) , where m is mass of water, C is it specific heat, h is the heat transfer co-efficient (mostly convective), T is temperature of water and t is time. Solving this simple differential equation with initial condition of T=378 at t =0 will provide a solution of the form
T= 298+80*e^(-hAt/(mC))
Note that I have ignored the effect of radiation. This is because, typically at these temperatures, the radiative heat losses are very small compared to convective heat losses. (The radiative heat loss term would have made it a more complicated differential equation)
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Srinidhi Kulkarni, Physics enthusiast
Answered Sep 24, 2016
Qualitatively answering, it's temperature will decrease with time from 378k to 298k in an exponential manner. This is because the evolution of temperature is given by
mC(dT/dt) = -hA(T-298) , where m is mass of water, C is it specific heat, h is the heat transfer co-efficient (mostly convective), T is temperature of water and t is time. Solving this simple differential equation with initial condition of T=378 at t =0 will provide a solution of the form
T= 298+80*e^(-hAt/(mC))
Note that I have ignored the effect of radiation. This is because, typically at these temperatures, the radiative heat losses are very small compared to convective heat losses. (The radiative heat loss term would have made it a more complicated differential equation)
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the energy lost once cannot be renewed
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