Math  /  Algebra

QuestionA water sample absorbed 2,851 J from metal. The 42.6 g metal cooled from 209°C to 40°C. Find the metal's specific heat. Cmetal=[?]JgC C_{\text{metal}} = [?] \frac{\mathrm{J}}{\mathrm{g} \cdot{ }^{\circ} \mathrm{C}} Remember, qmetal=qH2Oq_{\text{metal}} = -q_{\text{H}_2\mathrm{O}}.

Studdy Solution
Calculate the specific heat of the metal.
Cmetal=2,851 J42.6 g×169C=0.397JgCC_{metal} = \frac{2,851 \mathrm{~J}}{42.6 \mathrm{~g} \times169^{\circ} \mathrm{C}} =0.397 \frac{\mathrm{J}}{\mathrm{g} \cdot{ }^{\circ} \mathrm{C}}The specific heat of the metal is 0.397JgC0.397 \frac{\mathrm{J}}{\mathrm{g} \cdot{ }^{\circ} \mathrm{C}}.

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