A gold ring that weighing 3.81 g is heated to 84.0∘Cand placed in 50.0g of H2O at 22.1∘C. What is the final temperature?Specific Heat CapacityJ⋅g−1⋅C−1Au0.129H2O4.18

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A gold ring that weighing 3.81 g is heated to $84 {.0}^{\circ} C$ and placed in $50.0 g$ of $H_{2} O$ at ${22.1}^{\circ} C$ . What is the final temperature?

Specific Heat Capacity	$(J \cdot g^{- 1} \cdot C^{- 1})$
Au	0.129
$H_{2} O$	4.18

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22.2° C

24.0° C

26.5° C

53.1° C

answer is A.

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Detailed Solution

$\begin{array}{l} 3.81 \times 0.129 \times (84 - T) = 50 \times 4.18 (T - 22.1) \\ 84 - T = 425 (T - 22.1) \\ \Rightarrow T = {22.2}^{\circ} C \end{array}$