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When a substance having mass 3 kg receives 600 cal of heat, its temperature increases by 10 °C. What is the specific heat of the substance? |
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Answer» Data: m = 3 kg = 3000 g, Q = 600 cal, ΔT = 10 °C, c = ? Q = mcΔT ∴ C = \(\cfrac{Q}{mΔT}\) = \(\cfrac{600\,cal}{3000\,g\times10^\circ C}\) = 0.02 cal/(g.°C) This is the specific heat of the substance. |
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