An unknown metal of mass $ 192 g $ heated to a temperature of $ 100^{\circ} \mathrm{C} $ was immersed into a brass calorimeter of mass $ 128 \mathrm{~g} $ containing $ 240 \mathrm{~g} $ of water at a temperature of $ 8.4^{\circ} \mathrm{C} $. Calculate the specific heat of the unknown metal if water temperature stabilizes at $ 21.5^{\circ} \mathrm{C} . $ ( Specific heat of brass is $ 394 \mathrm{~J} \mathrm{~kg}^{-1} \mathrm{~K}^{-1} $ )

Question

An unknown metal of mass $ 192 g $ heated to a temperature of $ 100^{\circ} \mathrm{C} $ was immersed into a brass calorimeter of mass $ 128 \mathrm{~g} $ containing $ 240 \mathrm{~g} $ of water at a temperature of $ 8.4^{\circ} \mathrm{C} $. Calculate the specific heat of the unknown metal if water temperature stabilizes at $ 21.5^{\circ} \mathrm{C} . $ ( Specific heat of brass is $ 394 \mathrm{~J} \mathrm{~kg}^{-1} \mathrm{~K}^{-1} $ ), $ 916 \mathrm{~J} \mathrm{~kg}^{-1} \mathrm{~K}^{-1} $, $ 458 \mathrm{Jkg}^{-1} \mathrm{~K}^{-1} $, $ 654 \mathrm{Jkg}^{-1} \mathrm{~K}^{-1} $, $ 1232 \mathrm{Jkg}^{-1} \mathrm{~K}^{-1} $

MARKS App · Accepted Answer

Conservation of energy: m s Δ T = m b s b + m w s w Δ T w 192 × 10 - 3 s 100 - 21.5 = 128 × 394 + 240 × 4182 × 10 - 3 × 21.5 - 8.4 s = 50432 + 1003680 13.1 15072 = 916 . 19 ≃ 916 J k g - 1 K - 1

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