The circuit below is used to heat water kept | vedclass

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(C) In the given condition, the net heat gained by the water is the difference between the heat generated by the heater and the heat lost according to

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(C) In the given condition, the net heat gained by the water is the difference between the heat generated by the heater and the heat lost according to Newton's law of cooling.Let $i$ be the current, $R_{1}$ be the resistance of the heater, $m$ be the mass of water, $S$ be the specific heat capacity, $T$ be the temperature of water, $T_{0}$ be the ambient temperature, and $K$ be a constant.The rate of heat gain is given by:$i^{2} R_{1} - K(T - T_{0}) = m S \left(\frac{dT}{dt}\right)$Rearranging the terms to form a differential equation:$\frac{dT}{dt} = \frac{i^{2} R_{1}}{m S} - \frac{K}{m S}(T - T_{0})$$\frac{dT}{dt} = \left(\frac{i^{2} R_{1} + K T_{0}}{m S}\right) - \frac{K}{m S} T$Let $A = \frac{i^{2} R_{1} + K T_{0}}{m S}$ and $B = -\frac{K}{m S}$. Then:$\frac{dT}{dt} = A + BT$Integrating this equation leads to an exponential relationship between temperature $T$ and time $t$ of the form:$T(t) = T_{final} - (T_{final} - T_{initial}) e^{-kt}$This represents a curve that starts at the initial temperature and asymptotically approaches a steady-state temperature. This behavior is correctly depicted by graph $C$.

The circuit below is used to heat water kept in a bucket. Assuming heat loss only by Newton's law of cooling, the variation in the temperature of the water in the bucket as a function of time is depicted by

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