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A heating element of mass $100 \mathrm{~g}$ and having specific heat of $1 \mathrm{~J} /\left(\mathrm{g}^{\circ} \mathrm{C}\right)$ is exposed to surrounding air at $27^{\circ} \mathrm{C}$. The element attains a steady state temperature of $127^{\circ} \mathrm{C}$, while absorbing 100W of electric power. If the power is switched Off, then approximate time taken by the element to cool down to $126^{\circ} \mathrm{C}$ will be (neglect radiation)
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The correct answer is:
$1.0 \mathrm{~s}$
Given, power absorbed by element, $P=100 \mathrm{~W}$
Mass of element, $m=100 \mathrm{~g}=0.1 \mathrm{~kg}$
Change in temperature,
$\Delta T=(127-126)^{\circ} \mathrm{C}=1{ }^{\circ} \mathrm{C}$
Specific heat of heating element,
$s=1 \mathrm{~J} /\left(\mathrm{g}^{\circ} \mathrm{C}\right)=1 \times 10^3 \mathrm{~J} /\left(\mathrm{kg}^{\circ} \mathrm{C}\right)$
$\therefore$ Heat rejected to cool from $127^{\circ} \mathrm{C}$ to $126^{\circ} \mathrm{C}$, $Q=m s \Delta T=0.1 \times 1 \times 10^3 \times 1=100 \mathrm{~J}$
$\therefore$ Time taken $=\frac{Q}{P}=\frac{100}{100}=1 \mathrm{~s}$
Mass of element, $m=100 \mathrm{~g}=0.1 \mathrm{~kg}$
Change in temperature,
$\Delta T=(127-126)^{\circ} \mathrm{C}=1{ }^{\circ} \mathrm{C}$
Specific heat of heating element,
$s=1 \mathrm{~J} /\left(\mathrm{g}^{\circ} \mathrm{C}\right)=1 \times 10^3 \mathrm{~J} /\left(\mathrm{kg}^{\circ} \mathrm{C}\right)$
$\therefore$ Heat rejected to cool from $127^{\circ} \mathrm{C}$ to $126^{\circ} \mathrm{C}$, $Q=m s \Delta T=0.1 \times 1 \times 10^3 \times 1=100 \mathrm{~J}$
$\therefore$ Time taken $=\frac{Q}{P}=\frac{100}{100}=1 \mathrm{~s}$
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