A 192 text g unknown metal at 100^circtext C is dropped into a brass calorimeter (128 text g ) containing 240 text g of water at 8.4^circtext C . The final equilibrium temperature is 21.5^circtext C . Calculate the specific heat of the metal (Specific heat of brass = 394 text J kg ^ -1 text K ^ -1 , water = 4184 text J kg ^ -1 text K ^ -1 ).

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EEJ MAIN Physics QUESTION #7055

Question 1

A $192 \text{ g}$ unknown metal at $100^\circ\text{C}$ is dropped into a brass calorimeter ($128 \text{ g}$) containing $240 \text{ g}$ of water at $8.4^\circ\text{C}$. The final equilibrium temperature is $21.5^\circ\text{C}$. Calculate the specific heat of the metal (Specific heat of brass $= 394 \text{ J kg}^{-1}\text{K}^{-1}$, water $= 4184 \text{ J kg}^{-1}\text{K}^{-1}$).

$458 \text{ J kg}^{-1}\text{K}^{-1}$
$1232 \text{ J kg}^{-1}\text{K}^{-1}$
$916 \text{ J kg}^{-1}\text{K}^{-1}$✔️
$654 \text{ J kg}^{-1}\text{K}^{-1}$

Correct Answer Explanation

Using the Principle of Calorimetry: Heat lost by metal = Heat gained by (water + calorimeter).

$m_m s_m (100 - 21.5) = (m_w s_w + m_c s_c) (21.5 - 8.4)$

$0.192 \times s_m \times 78.5 = (0.24 \times 4184 + 0.128 \times 394) \times 13.1$

$15.072 s_m = (1004.16 + 50.432) \times 13.1 = 1054.592 \times 13.1 = 13815.15$

$s_m \approx 916.6 \text{ J kg}^{-1}\text{K}^{-1}$

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