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Calorimetry

When 300 J of...

Question

When 300 J of heat is added to 25 g of sample of a material, its temperatures rises from $25^{\circ} C$ to $45^{\circ} C$ . Calculate its thermal capacity.

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Solution

Heat $Q = m S Δ T$ where $S =$ specific heat of the material.
So, $300 = (25 / 1000) S (45 - 25)$ or $S = 600 J / k g^{o} C$
Thermal capacity, $c = m S = (25 / 1000) (600) = 15 J /^{o} C$

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