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A solid maintained at $t_{1}^{\circ} \mathrm{C}$ is kept in an evacuated chamber at temperature $t_{2}^{\circ} \mathrm{C}$ $\left(t_{2} \gg t_{1}\right) .$ The rate of heat absorbed by the body is proportional to
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Verified Answer
The correct answer is:
$t_{2}-t_{1}$
As we know that, the rate of cooling is
$$
\frac{d \theta}{d t}=b A\left(\theta-\theta_{0}\right)
$$
where, $b A=$ constant $\Rightarrow \quad \frac{d \theta}{d t} \propto \theta-\theta_{0}$
Also $-\frac{d T}{d t} \propto T-T_{0}$
where, $T=$ temperature and $\quad t=$ time
So, the rate of heat absorbed by cooled body is proportional to $t_{2}-t_{1}$
$$
\frac{d \theta}{d t}=b A\left(\theta-\theta_{0}\right)
$$
where, $b A=$ constant $\Rightarrow \quad \frac{d \theta}{d t} \propto \theta-\theta_{0}$
Also $-\frac{d T}{d t} \propto T-T_{0}$
where, $T=$ temperature and $\quad t=$ time
So, the rate of heat absorbed by cooled body is proportional to $t_{2}-t_{1}$
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