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(D) The heat supplied to the water is given by $dQ = m s dT$,where $m$ is the mass and $s$ is the specific heat capacity.Given $s = 1 \ J \ g^{-1} \ K

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$m \ln \left(\frac{T_2}{T_1}\right)$

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(D) The heat supplied to the water is given by $dQ = m s dT$,where $m$ is the mass and $s$ is the specific heat capacity.Given $s = 1 \ J \ g^{-1} \ K^{-1}$.The change in entropy $dS$ is defined as $dS = \frac{dQ}{T}$.Substituting $dQ$,we get $dS = \frac{m s dT}{T}$.To find the total change in entropy $\Delta S$,we integrate from $T_1$ to $T_2$:$\Delta S = \int_{T_1}^{T_2} \frac{m s dT}{T} = m s \int_{T_1}^{T_2} \frac{dT}{T}$.$\Delta S = m s \ln \left(\frac{T_2}{T_1}\right)$.Since $s = 1$,the change in entropy is $\Delta S = m \ln \left(\frac{T_2}{T_1}\right)$.

Water of mass $m$ gram is slowly heated to increase the temperature from $T_1$ to $T_2$. The change in entropy of the water,given specific heat of water is $1 \ J \ g^{-1} \ K^{-1}$,is

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