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A solid of $2 \mathrm{~kg}$ mass absorbs $50 \mathrm{~kJ}$ when its temperature is raised from $20^{\circ} \mathrm{C}$ to $70^{\circ} \mathrm{C}$. The specific heat capacity of this solid in unit of $\mathrm{J} / \mathrm{kg}{ }^{\circ} \mathrm{C}$ is
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The correct answer is:
500
Heat absorbed by a solid is given by
$\Delta Q=m s \Delta T$
where, $s=$ specific heat of solid.
$\therefore \quad s=\frac{\Delta Q}{m \cdot \Delta T}=\frac{50 \times 1000}{2 \times\left(70^{\circ}-20^{\circ}\right) \mathrm{C}}$ $[\because \Delta Q=50 \mathrm{~kJ}=50 \times 1000 \mathrm{~J}]$
$=500 \mathrm{~J} \mathrm{~kg}^{-1 \circ} \mathrm{C}^{-1}$
$\Delta Q=m s \Delta T$
where, $s=$ specific heat of solid.
$\therefore \quad s=\frac{\Delta Q}{m \cdot \Delta T}=\frac{50 \times 1000}{2 \times\left(70^{\circ}-20^{\circ}\right) \mathrm{C}}$ $[\because \Delta Q=50 \mathrm{~kJ}=50 \times 1000 \mathrm{~J}]$
$=500 \mathrm{~J} \mathrm{~kg}^{-1 \circ} \mathrm{C}^{-1}$
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