The circuit below is used to heat water kept | vedclass

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Question

$(c)$ In given condition,

Heat gained by water heater - Heat lost governed by Newton's law of cooling = Net heat gained by water

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Vedclass · Accepted Answer

Vedclass · Answer

$(c)$ In given condition,

Heat gained by water heater - Heat lost governed by Newton's law of cooling = Net heat gained by water

$\Rightarrow i^{2} R_{1}-K\left(T-T_{0}ight)=m S\left(\frac{\Delta T}{\Delta t}ight)$

Here, $T=$ temperature and $t=$ time.

$\Rightarrow \frac{d T}{d t} =\frac{i^{2} R_{1}}{m S}-\frac{K}{m S}\left(T-T_{0}ight)$

$=\frac{i^{2} R_{1}}{m S}+\frac{K T_{0}}{m S}-\frac{K}{m S} T$

Above equation can be written as,

$\frac{d T}{d t}=A+B T$

$	ext { or } \frac{d T}{A+B T}=d t$

Integrating, we get

$\Rightarrow \frac{1}{B} \cdot \log _{e}(A+B T)=t$

$\Rightarrow A+B T=e^{t B}$

$\Rightarrow T=\frac{1}{B}\left(e^{t B}-Aight)$

Hence, temperature $T$ depends on time $t$ exponentially.

$	herefore$ Best suitable graph is

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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